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Stewart III DE-238 - History

Stewart III DE-238 - History



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Stewart III

(DE-238: dp. 1200, 1. 306'0 b. 36'10, dr. 12'3~, s. 19.5 k.; cpl. 2i6; a. 3 3N, 2 0mm., 8 20mm., 2 act., 8 dcp., 1 dcp. (hh.), 3 21" tt.; cl. Edsall)

The third Stewart (DE-238) was laid down at Houston, Tex., by the Brown Shipbuilding Co. on 15 July 1942; launched on 22 November 1942; sponsored by Mrs. William A. Porteos, Jr., and commissioned on 31 May 1943, Lt. Comdr. B. C. Turner, USNR, in command.

Stewart remained at Houston until 10 June, when she shifted to Galveston. She entered the drydock there on the 14th and exited on the 16th. The following day she got underway for New Orleans, La., where she reported for duty to the Commandant of the 8th Naval District and to the Commander, Operational Training Command, Atlantic Fleet (COTCLANT). The destroyer escort departed New Orleans on 22 June to conduct shakedown training in the vicinity of Bermuda; completed it a month later; and sailed for Philadelphia. After six days at the Philadelphia Navy Yard, Stewart headed south to Miami, Fla., from whence she operated, conducting patrols and exercises, until 29 October. She put to sea; headed north, and, on the 31st, arrived at Norfolk, Va.

After a cruise up the Potomac River, during which she visited Quantico, Va., and the Washington Navy Yard, Stewart commenced a tour of duty training prospective destroyer-escort crews out of Norfolk. That assignment continued for the next three and one-half months, broken only by two temporary assignments escorting convoys from Tompkinsville, N.Y., to the Virginia Capes area. On 17 March 1944 she sailed from Norfolk for Tompkinsville; arrived there the next day; and put to sea, on the 19th, in the screen of a convoy bound-via Argentia, Newfoundland-for Reykjavik, Iceland. She returned to Tompkinsville on 10 April and sailed for Norfolk on the 12th. She arrived there on the 13th, drydocked from the 14th to the 16th, and joined the screen of convoy Task Unit (TU) 29.6.1 on 25 April.

Stewart sailed with her convoy via Aruba in the Netherlands West Indies and made Cristobal in the Canal Zone on 3 May. The following day, she put to sea with the convoy and escorted it as far as Guantanamo Bay, Cuba. There, she parted company with the other ships and steamed independently to Bermuda.

The destroyer escort arrived at Port Royal on 10 May and, for the next week, made experimental attacks on the captured Italian submarine Rea. From the 18th to the 23d, Stewart participated in a search off Bermuda for an unidentified radio direction finder contact. She made one depth charge attack on the 18th, but the results were inconclusive. On the 23d, she put back into Port Royal and remained there four days.

Stewart departed Port Royal again on the 27th, this time in a hunter-killer group composed of Rhind (DD-404) and Wainwright (DD-419), in addition to herself. On 3 June, the three warships rendezvoused with convoy UC 24, and the group sailed north. Stewart was detached on the 8th and on the 9th, put into Boston, Mass. On the 25 she sailed to Casco Bay, Maine, and the following day, conducted antisubmarine warfare (ASW) exercises with the captured Italian submarine Vortice. On the 27th, she sailed south to Norfolk. Stewart arrived on the 29th and put to sea again on 1 July in the escort of convoy UGF 12. The destroyer escort screened the convoy to Naples, Italy, where it arrived on 15 July. She departed Naples on 21 July in the screen of the return convoy, GUF 12, and moored at the Brooklyn Navy Yard on 3 August.

In mid-August, she returned to Casco Bay for two days of training; then she entered drydock at Boston on the 17th. She was refloated on the 21st and soon got underway to join another convoy at Norfolk. Stewart arrived at Norfolk on 22 August. On the 24th, she began another voyage to Naples, returning to the United States at New York on 26 September. When she shifted to Casco Bay on 9 October, she took up ASW training again with Vortice. On 20 October, she returned to Boston, from whence she sailed two days later in the screen of convoy CU 44. On that same day, Stewart dropped four depth charges at a sound contact but had to abandon the search and rejoin the convoy. She entered the River Clyde and moored there on 2 November. Eight days later, the destroyer escort sailed for the Unite States and arrived in New York on 22 November.

Following another round of ASW training off Nantucket Island, this time with the Italian submarine Manelz, she departed Boston on 10 December in the screen of another convoy. Ten days later, she entered Plymouth Sound. On the night of 23 and 24 December she shifted to the Isle of Wight where she joined another convoy getting underway for America. Between January and June 1945, Stewart escorted three more convoys to England, one to Falmouth and two to Liverpool. Between each round-trip voyage, she trained off the New England coast. On the return voyage from the second of these missions, Stewart was called upon to assist SS Saint Mihiel in fighting fires caused by a collision with SS Nashbulk. Following her final voyage to England, Stewart put into the New York Navy Yard for 18 days of availability. On 24 June 1945, she departed New York for Norfolk, arriving there on the 26th. After a breif stop, she continued on to Guantanamo Bay, Cuba, where she conducted training exercises from 30 June until 12 July.

She cleared the area on the 12th in company with Edsall (DE-129) and Moore (DE-240). The three warships transited the Panama Canal on 16 July and made San Diego on the 24th. Four days at the Naval Repair Base followed; then Wilhoite (DE-397) joined Stewart and the other two destroyer escorts as they headed for Pearl Harbor on the 28th. The formation reached Pearl Harbor on 4 August, and Stewart conducted training, first with Spearfish (SS-190), then with Baltimore (CA-68) until 5 September when she departed for the west coast. She stopped at San Diego from 11 to 13 September; then continued on to the Canal Zone.

She retransited the canal on 22 September and made Philadelphia on the 27th. One month later, Stewart reported for duty to the Atlantic Reserve Fleet at Philadelphia. Stewart was placed out of commission, in reserve, in January 1947 and berthed in Florida. Stewart changed berthing areas three times between 1947 and 1969—first to Charleston in 1958, then to Norfolk in 1959, and finally to Orange, Tex., in 1969. In 1972, the destroyer escort underwent inspection and survey and was found to be unfit for further naval service. Consequently, her name was struck from the Navy list on 1 October 1972. On 25 June 1974, she was donated to the state of Texas, and she now stands landlocked next to Cavalla (SS 244) at Pelican Island in Galveston, Tex.


Our History

The history of Lime Instruments starts in 2006, when C. Jim Stewart III emerges from retirement to begin a new era in the Houston oil industry. Prior to 2006, “Jimmy” Stewart, 58, spent 34 years of full-time employment with Stewart & Stevenson, the family business that his great-grandfather, C. Jim Stewart, started in 1902 as a blacksmith shop in downtown Houston, Texas. Over the past century, Stewart & Stevenson has become an icon in the Houston oil industry and has successfully transformed horseshoes into horsepower by emerging as a major distributor/packager of diesel engines and other products. Jimmy ultimately became an executive vice president of Stewart & Stevenson and sat on the company’s board before his retirement at the age of 54.

In 2006, Jimmy decided to get back into industry by acquiring Supreme Electrical Service & Supply Co. Inc. (SES), a privately held oilfield service company located in Houston, Texas. SES, a 58-employee company founded in 1992, provided electrical rig-up services to drilling rig manufacturers. SES gave Jimmy a talented employee base as well as a business foundation to diversify into new product offerings and services. In 2006, Jimmy brought on board his son, Robert Ross Stewart Sr, who also worked at Stewart & Stevenson for over 10 years. While at Stewart & Stevenson, “Rob” Stewart worked in a business development role for the company’s oilfield control systems group. In March 2007, shortly after joining his father at SES, Rob quickly worked to assemble a team to start the Controls & Instrumentation division of SES. In 2011, the Controls & Instrumentation division was branded as Lime Instruments. In 2012, the Houston Business Journal (HBJ) deemed Lime Instruments to be the fourth fastest growing business in the City of Houston, and awarded Lime Instruments the title ‘Enterprise Champion’ in recognition of 613.4% market growth from FY2009 – FY2012.


Let's put the vote-by-mail 'fraud' myth to rest

Widespread calls to conduct the 2020 elections by mail, to protect voters from COVID-19 exposure, are being met with charges that the system inevitably would lead to massive voter fraud. This is simply not true.

Vote fraud in the United States is exceedingly rare, with mailed ballots and otherwise. Over the past 20 years, about 250 million votes have been cast by a mail ballot nationally. The Heritage Foundation maintains an online database of election fraud cases in the United States and reports that there have been just over 1,200 cases of vote fraud of all forms, resulting in 1,100 criminal convictions, over the past 20 years. Of these, 204 involved the fraudulent use of absentee ballots 143 resulted in criminal convictions.

Let’s put that data in perspective.

One hundred forty-three cases of fraud using mailed ballots over the course of 20 years comes out to seven to eight cases per year, nationally. It also means that across the 50 states, there has been an average of three cases per state over the 20-year span. That is just one case per state every six or seven years. We are talking about an occurrence that translates to about 0.00006 percent of total votes cast.

Oregon is the state that started mailing ballots to all voters in 2000 and has worked diligently to put in place stringent security measures, as well as strict punishments for those who would tamper with a mailed ballot. For that state, the following numbers apply: With well over 50 million ballots cast, there have been only two fraud cases verifiable enough to result in convictions for mail-ballot fraud in 20 years. That is 0.000004 percent — about five times less likely than getting hit by lightning in the United States.

This hardly seems like a world in which “thousands and thousands of people [are] sitting in somebody’s living room, signing ballots all over the place.”

We should make two things clear. First, there is no excuse for any type of voter or election fraud, by any method. States are justified in creating systems that are intended to deter and detect fraud, and for prosecuting it when discovered. All do.

Voting by mail presents challenges to the prevention of voter fraud that voting in person lacks. Most obviously, in-person voting occurs in public. A voter must announce their name out loud, and it is checked against the voter registration list. All states make provisions for some form of objectors, who can question the identity of the person at the check-in table, within the constraints of state law. Some states require a photo ID to be shown. Many states require the voter to sign a poll book. These and other procedures have been in place for a century-and-a-half, since the widespread election reforms of the 1880s and 1890s.

Second, no voting methodology is perfect. In-person voting has its own examples of fraud, however rare. It is also full of stories of missing power cords, missing keys, an inadequate number of ballots, machines that switched the voter’s intent, improper application of ID requirements, long lines and more. Nonetheless, in-person voting also has a role to play even in states that use the 100 percent mail-ballot election model.

As with in-person voting, states have methods to guard against fraudulently casting votes by mail too. Most have signature-matching requirements, either to scrutinize the application, the returned ballot, or both. We have seen this done effectively using a mix of human oversight and technology. Many states restrict who can return a ballot for a voter, or require those who return ballots on the behalf of others to identify themselves on the return envelope. Finally, the states with the most expansive vote-by-mail systems — such as Colorado, Oregon, Utah and Washington — send ballots to all registered voters and rely on the steady stream of mail between election offices and voters to keep the rolls clean, and to minimize the number of stray ballots that might be distributed.

Expanding voting by mail will be a challenge in most states in 2020. Logistical and security issues will need to be reviewed to ensure that every registered voter can do so safely and effectively, and that no one votes more than once. But we reiterate: There is no evidence that mail-balloting results in rampant voter fraud, nor that election officials lack the knowledge about how to protect against abuses.

Amber McReynolds is CEO of the National Vote at Home Institute. She formerly was the director of elections for Denver, where she helped to design and implement Colorado’s vote-at-home system. Follow her on Twitter @AmberMcReynolds.

Charles Stewart III is the Kenan Sahin Distinguished Professor of Political Science at MIT, the director of the MIT Election Data and Science Lab, and co-director of the Caltech/MIT Voting Technology Project. Follow him on Twitter @cstewartiii.


Issues

Political positions

Gun rights

Mills is a strong advocate for gun rights, and is well known for a YouTube video from January 2013 in which he spoke out against Rep. Rick Nolan's position on gun control. Nolan, the current representative from Mills' district, wanted to crack down on assault weapons in response to the Sandy Hook Elementary School shooting. Mills responded with a video in which he compared the damage done by a shotgun to the damage done by a Huldra AR-15, saying that a shotgun was actually more destructive, and that limiting Second Amendment rights was not the proper solution to prevent school shootings. ⎗]

Health care

Mills said that he opposed the Affordable Care Act and would replace it with a free-market system. ⎘]

Same-sex marriage

Mills has not taken a position on same-sex marriage because "the state has already decided the issue." ⎙]

Comments on ISIS

During the 2014 congressional elections, U.S. military involvement in the Middle East became an important topic of debate among congressional candidates. Unlike many U.S. congressmen from both parties, Rep. Rick Nolan (D), Mills' opponent, warned against taking military action against the Islamic State in Iraq and Syria (ISIS). Nolan urged President Barack Obama not to make a unilateral decision to mobilize the military without the consent of Congress. On August 29, 2014, Nolan stated, "American blood and treasury should not be made without the full consideration by all 535 members of the Congress of the United States." ⎚] He added, "When we get ourselves involved in that conflict, then we become a part of the problem and the solution becomes ours." ⎚]

Shortly after making these comments, Nolan issued a statement regarding ISIS. Nolan wrote, "I encourage them to employ the same intelligence resources – and the same selective, highly effective means they used to bring down Osama Bin Laden. Special operations of this kind do not involve U.S. troops on the ground, the killing of innocent people, or the re-involvement of the United States in another terribly destructive, expensive, open-ended conflict in that region.” ⎛]

Mills disagreed with Nolan, stating, "Like it or not we have to play some role here. And for Congressman Nolan to take those positions, I think, leaves America in an even weaker position than we have been in the past. We can't lead from behind. We have to lead." ⎜]

According to a 2014 poll published in The Washington Post, the majority of voters' views aligned more closely with Mills' than Nolan's on the issue of taking action in the Middle East. About 91 percent of voters believed ISIS to be a serious threat, and 71 percent supported the U.S. government ordering air strikes against the Sunni insurgents in Iraq. ⎝]

Campaign themes

Mills listed the following issues on his campaign website: ⎞]

  • Grow the economy from Main Street Up, not Wall Street Down: The Eighth District is a Main Street economy and job growth here comes from the ground up. That means we need tax reform that’s geared toward small business growth. Stewart doesn’t believe that Washington creates jobs- entrepreneurs and business owners create jobs.
  • Natural Resources: Stewart believes that one way to increase employment here in Minnesota’s Eighth District is to have federal policies which will allow us to fully capitalize on our mineral and timber resources. It’s critical that Washington stop blocking our ability to extract natural resources. That’s why Stewart supports the development of the Polymet Project.
  • Respect for Our Constitutional Rights: As a hunter and a competitive shooter, Stewart is staunch supporter of the Second Amendment, but he also believes that the entire Constitution must be protected and upheld in Washington.
  • Healthcare Reform: Stewart believes that Obamacare has failed in achieving its stated goal- it set out to insure the uninsured, but it has resulted in “uninsuring” the previously insured. With his background in administering Fleet Farm’s self managed health plans, Stewart knows how to implement market based healthcare reforms that will increase the supply of medical services, reduce the demand through prevention and actually make the healthcare delivery more efficient.
  • Cut Wasteful Spending and Rein in the National Debt: Our nation is more than $17 trillion in debt and politicians in Washington keep kicking the can down the road. We are quickly approaching the point where the annual debt servicing payments will grow to unsustainable levels. Our debt is endangering our children’s future and threatening our ability to leave this nation in better shape than we found it.

Media portrayal

Mills' long hair caused a variety of responses from both the media and voters. Whereas some saw his hair as making him look youthful and attractive, others thought that it made him look immature and unprofessional. Regardless, it won him acclaim as the "Brad Pitt of the Republican Party." ⎠] Mills was not concerned by the criticism, and embraced the differences between him and other candidates. In one interview, he compared himself with the typical candidate, stating, "I don’t look the part. I don’t shave every day, I have long hair, I don’t necessarily want to wear a suit unless I have to. I don’t feel comfortable with cookie-cutter molds of what is considered a traditional candidate." While this image could have helped him in his bid against incumbent Rick Nolan, Mills also ran the risk of voters focusing more on his appearance than the political message of his campaign. His solution was to focus his energy on a serious campaign, while letting others "promote the novelty of it." ⎠]


About us

18 Apr 2021: The question has been asked what proportion of men named Stewart, Stuart or a variant of the name are royal Stewarts. Currently, in the Stewart DNA Project, Royal Stewarts can be detected from the results of their Y-37 DNA tests. 739 men with the surname Stewart or a close variant have tested for 37 Y markers. Of these 218 are Royal Stewart descendants. So for the Stewart DNA project, 29.5 % of men named Stewart or a variant are descended from the Royal Stewart line.

24 July 2018. We recently identified a small subgroup of Ancient Stewarts who descend, in all likelihood, from Walter Stewart, Earl of Menteith. He was the son of Walter Stewart, 3rd High Steward, and brother of Alexander Stewart, 4th High Steward. These testers are in the subgroup R-L745>Y138948. This is the third subfamily which lies below the R-L744, L745, L746 group. Descendants of James, 5th High Steward are in R-L745>Z38845, and descendants of John Stewart of Bonkyl are in R-L745>S781.

1 Apri1 2016. Marker for male line descendants of King Robert III (1337-1406):
The results of the Big Y test for a documented male line descendant of Sir John Stewart of Blackhall & Ardgowan, d. c.1412, an illegitimate son of King Robert III, have now been received and analysed. Alex Williamson, author of The Big Tree http://www.ytree.net/ has identified a new SNP carried by this individual, which has been given the name ZZ52. Our two other Big Y test results, that is one for Earl Castle Stewart, a descendant of Robert Stewart, Duke of Albany, and the other for a documented descendant of Sir John Stewart, Sheriff of Bute, do not carry this SNP. Robert Stewart, Duke of Albany, and Sir John Stewart, Sheriff of Bute are both brothers of King Robert III. This means that ZZ52 must have occurred in Robert III or one of his male line descendants. In other words, ZZ52 is a distinct marker identifying descendants of Robert III. He may also have descendants who do not carry this, since the generation in which the mutation took place is not yet known.

20 July 2015 . Grouping within the Ancient Stewart family (i.e.descendants of Alexander Stewart, 4 th High Steward of Scotland) have been rearranged to be slightly more conservative than before. Grouping is now based on SNP results as analysed by Alex Williamson and shown in his websitefor Big-Y testers under R-P312. Results for the Royal Stewarts can be seen at http://www.ytree.net/DisplayTree.php?blockID=7 . The group is split into two (with subgroups below those levels) representing descendants of James, 5 th High Steward, characterized as R-746, and his younger brother, Sir John of Bonkyl, characterized as R-L746>S781. The two subclades have been split further, according to presence or absence of various SNPs found in Big-Y tests.

2 Jan 2015: As at 2 Jan 2015 there were 697 Y-DNA results in the Stewart DNA project, and 340 of those testers (or 49% of the total) did not have any matches within the project.

There were 243 testers who had at least one match within the project and they were listed in 61 different family groups, ranging in size from 114 down to 2. The largest single group was that of the Ancient Stewarts, with 114 testers in that group, or one of its subclades (that is, descendants from Alexander Stewart 4th High Steward of Scotland (1210-1283)). The next largest group was R1b Group Type 1c with 15 members, followed by I2b1 Group Type 1 with 11 members. In all there were 61 family groups and they accounted for 243 testers who had at least one match within the Stewart Project.

At that date there had been 46 Big-Y tests within the project and 277 mtDNA tests, of which 134 were for the Full Sequence and 215 for HVR1 & HVR2, and these tests were grouped into 14 different major haplogroups. We have also had 241 testers who have taken the Family Finder test.

24 Apr 2014: The SNP S781 can now be tested for at FTDNA and YSEQ, and results are coming through to split the Ancient Stewart family into two sub-families: Descendants of James Stewart, 5th High Steward of Scotland, are testing S781-, while descendants of Sir John Stewart of Bonkyl are testing S781+. Groupings within the Ancient Stewart family are now changing on a daily basis - see the Y-results page for details.

29 Jan 2014: The three Ancient Stewart SNPs (L744, L745, L746) can also be tested for at the ScotlandsDNA company where they are labelled S388, S463, S310. Now Dr Jim Wilson has discovered another SNP which arose in an early Stewart, Sir John Stewart of Bonkyl, ancestor of the Lennox Stuart lines and the English Stuart kings, as well as many other less noble families. This new SNP has been labelled S781, and a positive test for S781 indicates that the tester descends from Sir John of Bonkyl (1246-1298). Descendants of his elder brother James, 5th High Steward, ancestor of the Albany line of Stewart Kings of Scotland, have been found to test negative for S781, so this new SNP has the potential to divide the Ancient Stewarts into two different subfamilies, and there could be more Stewart SNPs to come.

8 Oct 2013: The odd result of one Ancient Stewart tester being found L745- but L744+, L746+, has been found to be an error. Rechecking his SNP results has revealed that he is L745+. This means that the three Ancient Stewart SNPs appear to be equivalent - a great relief to everyone.

30 June 2013: Recent analysis of the 111-marker Y-DNA results suggests that the L744+, L745-, L746+ tester (#75703) descends from one of the male-line Stewart ancestors, before they used the name Stewart. So #75703 could easily have come from a descendant of the old English line, i.e. from Alan Fitz Flaad, Lord of Oswestry, on the Welsh Border, whose son Walter Fitz Alan travelled north from England to Scotland in 1124 and became the First High Steward of Scotland. Or he could be descended from one of the early High Stewards of Scotland. What is now apparent is that by the time of Alexander, 4 th High Steward (1214-1283), Alexander's genome and those of his descendants included the L745+ SNP. As tester #75703 is L745-, he still has the original value of that SNP, so he cannot be a descendant of Alexander. However he has L744+ and L746+ so he must be a descendant of one of Alexander's near ancestors, and he is now labelled as descendant of some precursor to the Ancient Stewarts.

15 March 2013: Recent research by Doug Stewart has established that his male-line ancestry can be traced back to the Stewarts of Jedworth. Doug has also found persuasive evidence that these Stewarts are not male-line descendants of Sir John of Bonkyl (as had been surmised by some 18th century historians) but come from a local line of Seneschals in Jedworth, whose name had been anglicized to Steward or Styward. This is the male line of the Stewart Earls of Galloway, and the Stewarts of Castlemilk. The relevant subgroup (I2b1 Group Type 1) has therefore been subtitled as "Descended from the Stewards of Jedworth, ancestors of the Earls of Galloway".

28 Feb 2013: Analysis of 111 marker tests for a number of Ancient Stewart descendants suggests that there are two main lines of descent from Alexander Stewart, 4th High Steward of Scotland. These are (1) from his eldest son James, 5th High Steward, and then via King Robert II of Scotland. (2) from his younger son, Sir John of Bonkyl, and then from various cadet lines, including the Earls of Lennox and the English Royal Stuarts. Consequently a number of family groups have been renamed, according to which basic line they descend from according to the 111 marker tests, while two new groups have been created for descendants of Robert II, and Sir John of Bonkyl.

Jan 2013: A new SNP has been discovered within the large Scottish subclade of R-L21 known as the Scots Modal Group, Scottish Cluster, R1b-Pict, or R1bSTR47SCOTS. This subclade has been found to test positive for the SNP L1335.

July 2012: The Ancient Stewart SNP, L744, can now be positioned on the Y-DNA Family Tree, immediately below another SNP labelled DF41, which in turn lies below the widespread British marker R-L21. The longhand description of R-DF41 was included in the latest version of the ISOGG Y-DNA Tree a few days ago, where it has been designated R1b1a2a1b3a9. If all goes to plan, the new designation of R-L744 will be R1b1a2a1b3a9a, but this is still subject to confirmation by the ISOGG Committee.

17 Apr 2012 : ScotlandsDNA announced that "The DNA of the Duke of Buccleuch was found to be an exact match of a descendant of Charles Stewart of Ardshiel, who fought at Culloden, both men descended from Alan, the Seneschal of Dol, a Breton aristocrat. His family came to Britain in 1066 with William the Conqueror and then made its way to Scotland to found the Stewart line." The Duke of Buccleuch had been SNP tested and found to be positive for L744 (=S388) and L745 (=S463).

27 Dec 2011: An article in The Scotsman revealed that the Ethnoancestry DNA company has evidence that around 16% of Scotsman with the surname Stewart carry the DNA marker S310, the same marker as undoubted descendants of descendants of James V, and James VI and I. It is a subtype of the widespread Celtic marker S145, also known as L21. SNP S310 is identical to one of the marker s which FTDNA has found in a number of Stewarts to date, namely L746.

4 Nov 2011: Updates on news items in September 2011.

1. The tester who has the non-modal value of 12 for 406S1 has tested positive for L744 and L746, though negative for L745, so is closely related to the Ancient Stewart Group.

2. All Stewarts tested for the L743 SNP have proved negative, so this hasn't advanced our knowledge of the Scots Modal variety.

3. All members of the Ancient Stewart who have tested for L746 have had POSITIVE results, so it looks as though L746 is a marker for the Ancient Stewarts. Results for L744 are similarly positive for all Ancient Stewarts tested to date, and for all but one L745 testers. The L745- tester also has the value of 406s1=12 so he can be expected to be a relative.

30 Sept 2011: New results have been found in the 'Walk Through the Y' (WTY) Project which could have important ramifications for two different groups of Stewart testers.

Firstly a new SNP labelled L743 has been found in a man who belongs to the Scots Modal variety of the R1b haplogroup. As there are a large number of Stewarts who are known to belong to this group on the basis of their STR results (plus others presently unassigned within R1b who might also belong there) it is likely that the L743 mutation will provide a SNP test for membership in the Scots Modal Group. More testing is needed to confirm this but we are optimistic as to the outcome. If this comes to fruition members who test positive for this SNP will move into a revised haplogroup, R-L743, probably to be described as R1b1a2a1a1b4u or similar.

Secondly, three new SNPs (L744, L745 and L746) have been discovered in a tester who belongs to the Ancient Stewart family group. If one or more of these SNPs turns out to be a label for all members of the group, that would provide us with a specific SNP test for Ancient Stewarts in future. Again, more testing is needed, so I'll report results as they come through. If this comes to fruition members who test positive for one of these SNPs will move into a revised haplogroup, probably to be described as something like R1b1a2a1a1b4v.

20 Sept 2011: Discovery of a tester who is a close match to several members of the Ancient Stewart Group yet has the unusual value of 12 for 406S1, suggests that the definition of the Ancient Stewart signature should be extended to: GataH4=10, 406S1=11 or 12, 565=11. CONFIRMED.

14 June 2011 : A reanalysis of 67-marker results, combined with improved knowledge of some family relationships within the line of the Lennox Stuarts, has cast doubt on the usefulness of markers CDYa, b, and DYS464a, b, c, d as indicators of family lines within the Ancient Stewarts (see 20 Nov 2010 item 2, above).

The variation DYS464=14-15-16-17 appears to have arisen at least twice among Ancient Stewart descendants, so although it appears to mark the English Royal Stuarts, as all known descendants of that line have the 14-15-16-17 variation, it is believed to have occurred somewhere in the latter part of the Lennox/Darnley line, certainly since the time of Sir John Stuart, 3rd Earl of Lennox (c1490-1526). The Results section will be amended accordingly.

14 May 2011: A revised version of the Ancient Stewart DNA Relationship Tree, based on 67-marker DNA tests (Haplogroup R1b) from Stewart/Stuart descendants, was released. Details of Stewart relationships can be seen at http://www.chuckspeed.com/balquhidder/history/DNA_Stewarts%20May2011.htm

31 Mar 2011: The project now has 450 members.

20 Nov 2010 : Promising DNA indicators have been found for different branches of the Ancient Stewart line, descendants of Alexander Stewart, 4 th High Steward of Scotland. The tree relationship diagram and details of these indicators can be seen at http://www.chuckspeed.com/balquhidder/history/DNA_of_Ancient_Stewart_descendantsNov2010.htm

The first requirement for any descendant of the Ancient Stewarts is that he needs to be 10,11,11 for GataH4, 406S1 and 565. Then there are two further possibilities:

1. For most Ancient Stewart descendants from Alexander, 4 th High Steward of Scotland, certainly for the Scottish Royal Stewart line, and for the Stewarts of Appin and Ardsheal, he should have 14-15-17-17 for DYS464. There are several minor variations but this is the main 464 signature for the Ancient Stewarts.

2. For descendants from the English Royal Stuarts, and from their ancestral Scottish line back through the Lennox Stuarts, he should have 14-15-16-17 for DYS464. All men of this ancestry descend from Sir John Stewart of Bonkyle, second son of Alexander Stewart, 4 th High Steward of Scotland, though the DNA split must have occurred lower down the line than Sir John himself. That line produced an ancestor (identity unknown) of our friend Charles MacLeod-Stuart, and, later, a progenitor of the line of Henry Stuart, Lord Darnley, down through James I and VI, Charles II, and later descendants via noble but illegitimate lines.

The variation DYS=14-15-16-17 appears to apply to a very small subgroup of the Ancient Stewarts. It just happens to be the subgroup that contains the English descendants of Henry Darnley, heir to the Earl of Lennox, who married Mary Queen of Scots and whose son James I and VI inherited the English crown after the death of Queen Elizabeth I. King James V of Scotland would have had 14-15-17-17, but his grandson, James VI of Scotland (and I of England) inherited his 14-15-16-17 from his father Henry Darnley.

11 Nov 2010 : Belinda Dettmann was added to the Stewart Project as Co-Administrator. The project now has 415 members.

8 Nov 2010: A new Stuart Project has been created, catering for testers who belong to the family described therein as Royal Stewarts. The criterion for acceptance is that they must be within a short genetic distance of a particular DNA tester who is known to descend from King Charles II. Details can be found at http://www.familytreedna.com/public/stuart

7 Nov 2010 : Announcement of a new definition for the Y-DNA group of Royal and Ancient Stewarts. It requires 67 Y-DNA markers and applies to people in Haplogroup R1b1b2a1b5. Anyone in that subclade with a Y-DNA test with GATAH4=10, 406S1=11, and 565=11, is likely to descend from the line of the High Stewards of Scotland.

12 Sep 2010 : A revised version of the Ancient Stewart DNA Relationship Tree, based on 67-marker DNA tests (Haplogroup R1b) from Stewart/Stuart descendants, was released. Details of Stewart relationships can be seen at http://www.chuckspeed.com/balquhidder/history/R1b_Stewart_DNA_Aug2010.htm

30 Jan 2010 : Doug Stewart called our attention to his 67-marker test which belongs to Haplogroup I2b1. This is an important piece of information as Doug has traced his ancestry in an unbroken line back to William Stewart, 1st Laird of Dunduff, born circa 1500, Ambassador to France in the year 1528. That family traced its male line directly to King Robert II of Scotland via the Lords of Avondale. This line of ancestry challenges the idea that the Royal and Ancient Stewart line belonged to Haplogroup R1b. Research to elucidate the DNA signature of the Stewart High Stewards of Scotland is continuing, although the Y-DNA signature of the ancient Stewart line being similar to QHV9S is presently preferred.

18 Aug 2009 : A revised version of the Ancient Stewart DNA Relationship Tree, based on 67-marker DNA tests (Haplogroup R1b) from Stewart/Stuart descendants, was released. Details of Stewart relationships can be seen at
http://chuckspeed.com/balquhidder/history/R1b_Stewarts_67_Aug2009.doc

The haplotype designated ORIGIN is believed to be that of Alexander Stewart (1214-1283), 4th High Steward of Scotland. He was the ancestor of the Royal Stewart kings of Scotland (via King Robert II), and the Royal House of Stuart in the United Kingdom (via King James I and VI). He was also the progenitor of many other lines of Stewart or Stuart nobles and gentry in Scotland, England, Ireland and France. The ORIGIN haplotype does not represent the test result of a single person, but has been inferred from the pattern of Stewart descendants. It is entered in Ysearch as QHV9S.

7 Mar 2009 : Charles MacLeod-Stuart revealed that four members of the British aristocracy (two Dukes, a Lord and an Earl), all of royal Stewart ancestry, have had DNA tests results in the R1b haplogroup quite similar to his.

May 2008 : Dick Stewart loaded an Excel file at the Stewart-DNA list site and asked all Stewart DNA researchers to enter their ancestral details.

14 Jan 2008 : Belinda Dettmann put a diagram of Stewart R1b 37 marker relationships up on the web, now showing at http://www.chuckspeed.com/balquhidder/history/StewartYDNA.html

20 Nov 2006: Debi Stewart was added to the Project as third Project Administrator.

3 Nov 2006: The Stewart group now has 185 participants. Of the 171 kits returned there are 112 unique marker strings (results). This represents great diversity within the group. It means that 59 have matches within the group at the 25 marker level (those who have taken the 25 marker test). There are 85 unique 12 marker strings, 64 unique 37 marker and 11 with 67 markers.

16 July 2002 : The Stewart Project began with a test numbered 4137 from Kathi Bobb’s father, and with Kathi as Project Administrator.


The Stewart Companies:

Values Honored. Success Ensured.

Technology evolves, codes change, and economies come and go. But the foundational values that have driven The Stewart Companies’ success since the 1930s remain bedrock firm.

Our minds are keener, thanks to years of experience. Our work ethic is stronger, honed by years of competitive fire. And our drive to cultivate lasting relationships has never been more intense. After all, while anyone can land new customers, it’s the ability to develop and maintain repeat customers that truly reflects a company’s long-term value.

That is how Robert Stewart’s legacy lives on in his children, grandchildren and the company that bears his name.

Started in 1935 by:

Stewart III DE-238 - History

Story & Photos By Tom Behrens

Roberta Marie Christensen in her book, “Pioneers of West Galveston Island,” says the Stewart Mansion on Galveston Island left her in awe-struck silence when she was given permission to tour the building in 1988. “The once beautiful plaster walls, the architecture of the second floor balcony, the Spanish tile work and the four bathrooms — one with marble walls and the living room walls covered with vivid, larger than life murals of pirates. Guarding the entrance way was a huge leering pirate with a three-cornered hat and saber in hand. On the opposite wall was another pirate with a bandana and sword. Looking down from the balcony were five figures, the foremost armed with a machete.”

Before she went much further into her inspection of the once proud mansion, she received one admonishment from the caretaker, “There are ghosts in [this] house. My wife and I hear doors banging and noises in the middle of the night.”
Check the internet and other similar comments pop up: “Don’t go at night…it would be way too scary, especially the bathroom.” “Things go bump in the night.” There are even rumors that the family was killed and put into the walls of the mansion by Stewart himself before he committed suicide. Pretty creepy.
Is the Stewart family encased in the walls of the mansion? Did Maco Stewart Jr. kill his family and put them into the walls of the mansion before killing himself? History doesn’t back that story up. Maybe the teller of that tale read too many Edgar Allen Poe tales. Let’s take a brief excursion back in history about the Stewart Mansion and Ranch.
At one time, the mansion was the main house on the Stewart Ranch. In addition there were two houses for the ranch hands. In 1969, the ranch, minus the mansion, was given to the State of Texas, and became what is now Galveston Island State Park. “The ranch existed on both sides of the existing state highway,” says park superintendent Trey Goodman. “There are remnants that still exist of the old cattle ranch: dip tanks, wind mills and other things scattered around the park.”


House Of Tudor Genealogy Chart & Family Tree

It all began with her…. Margaret Beaufort, mother of Henry Tudor and descendant of King Edward III. Her descent was actually illegitimate, through Edward III’s son, John of Gaunt and his mistress Catherine Swynford. Though Gaunt later married Swynford and an act of government legitimized their children, they were expressly forbidden to inherit the throne by Gaunt’s legitimate son, King Henry IV. Margaret bore her only child, who became King Henry VII, at the age of thirteen. While the York and Lancaster branches of the royal family battled for the throne, her husband died and her brother-in-law fled to Brittany with the young Henry. His long exile ended on 15 August 1485 when King Richard III was defeated at Bosworth Field. The Plantagenet dynasty, having ruled England since 1154, ended in strife and the Tudor dynasty began.

Click here to view a .bmp (bitmap) image of a Tudor family tree. I apologize for the blurry text I’m searching out a better scan. Below is a more detailed text genealogy.

Henry Tudor, earl of Richmond later titled King Henry VII
son of Edmund Tudor and Margaret Beaufort
born 28 January 1457 at Pembroke Castle, Wales
claimed title of Henry VII, king of England c.1484
declared king 22 August 1485 at Battle of Redmoor Plain/Bosworth Field
coronation 30 October 1485 at Westminster Abbey, London
married to Elizabeth Plantagenet, princess of England, on 18 January 1486 at Westminster Abbey, London
Eight Children with Elizabeth Plantagenet:
Arthur, prince of Wales, born 20 September 1486
Margaret, born 28 November 1489
Henry VIII, king of England, born 28 June 1491
Elizabeth, born 2 July 1492 died 1495
Mary, born 18 March 1496
Edmund, duke of Somerset, born 21 February 1499 died 1500
Edward ?
Katherine, born and died 2 February 1503
died 22 April 1509 at Richmond Palace, Surrey
buried at the Henry VII ‘Lady Chapel’, Westminster Abbey, London

Princess Elizabeth Plantagenet, also called Elizabeth of York
daughter of Edward IV, king of England & Queen Elizabeth Woodville
born 11 February 1466 at Westminster Palace, London
married to Henry VII, king of England, on 18 January 1486 at Westminster Abbey, London

Eight Children
Arthur, prince of Wales, born 20 September 1486
Margaret, born 28 November 1489
Henry VIII, king of England, born 28 June 1491
Elizabeth, born 2 July 1492 died 1495
Mary, born 18 March 1496
Edmund, duke of Somerset, born 21 February 1499 died 1500
Edward ?
Katherine, born and died 2 February 1503

coronation 25 November 1487 at Westminster Abbey, London
died 11 February 1503 at Tower of London, London
buried at the Henry VII ‘Lady Chapel’, Westminster Abbey, London

Prince Arthur of Wales
born 20 September 1486 at St. Swithin’s Priory, Winchester
titled Prince of Wales 27 February 1490, Westminster Palace, London
married to Princess Katharine of Aragon on 14 November 1501 at St.Paul’s Cathedral, London
died 2 April 1502 at Ludlow Castle, Shropshire
buried at Worcester Cathedral

Princess Margaret Tudor later titled queen of Scots
born 28 November 1489 at Westminster Palace, London
married to James Stuart, titled King James IV of Scotland, 8 August 1503 at Holyrood Abbey, Edinburgh
titled Queen of Scotland 8 August 1503 at Holyrood Abbey, Edinburgh
Six Children with King James IV of Scotland:
James of Rothesay, duke of Rothesay, born 21 February 1507
daughter, born 15 July 1508
Arthur, duke of Rothesay, born 20 October 1509
James, later King James V of Scotland, born 15 April 1512
daughter, born November 1512
Alexander, duke of Ross, born 30 April 1514
widowed 1513
married to Archibald Douglas, 6th earl of Angus, on 4 August 1514 at Kinnoul Church
One Child with Archibald Douglas:
Margaret, Lady Douglas, born 1515
divorced 1528
married to Henry Stewart, Lord Methven I, on 3 March 1528
One Child with Henry Stewart:
Lady Dorothea Stewart, born ?

Go to the Scottish genealogy page.

Prince Henry Tudor, duke of York later titled King Henry VIII
born 28 June 1491 at Greenwich Palace, London
titled King of England 24 June 1509 at Westminster Abbey, London
married to Katharine of Aragon, princess of Spain, on 11 June 1509 at Grey Friars Church, Greenwich
Six Children with Katharine of Aragon:
daughter, born 31 January 1510
Henry (1), duke of Cornwall, born 1 January 1511
Henry (2), duke of Cornwall, born November 1513
son, born December 1514
Mary, later titled Mary I, queen of England, born 18 February 1516
daughter, born 10 November 1518
marriage annulled 1533
married to Anne Boleyn, marquess of Pembroke, on 25 January 1533 at Westminster, London
Three Children with Anne Boleyn:
Elizabeth I, queen of England, born 7 September 1533
Henry, duke of Cornwall, born 1534
son, born 29 January 1536
marriage annulled 1536
married to Jane Seymour on 20 May 1536
One Child with Jane Seymour:
Edward, later titled Edward VI, king of England, born 12 October 1537 at Hampton Court Palace
widowed 24 October 1537 at Hampton Court Palace
married to Anne of Cleves on 6 January 1540 at Greenwich Palace
marriage annulled 1540
married to Catherine Howard on 28 July 1540 at Hampton Court Palace
marriage annulled 1541
married to Katharine Parr on 12 July 1543 at Hampton Court Palace
died 28 January 1547 at Whitehall Palace, London
buried at St. George’s Chapel, Windsor Castle
mistresses
Elizabeth Stafford – no children
Elizabeth Blount, called Bessie Blount
One child with Elizabeth Blount:
Henry Fitzroy, duke of Richmond, born 1519

Go to the Stafford/Blount/Fitzroy genealogy page.

Princess Mary Tudor, queen of France and duchess of Suffolk
born 18 March 1496 at Richmond Palace, Surrey
married to King Louis XII of France on 9 October 1514 at Abbeville Cathedral, France
widowed 1514
married to Charles Brandon, duke of Suffolk, on 3 March 1515 in Paris

Three Children
Henry, earl of Lincoln, born 11 March 1515
Frances, born 16 July 1517
Eleanor, born 1519

died 25 June 1533 at Westhorpe Hall, Suffolk
buried at St. Mary’s Church, Bury St. Edmunds

Katharine of Aragon, princess of Spain
last child of Ferdinand of Aragon, king of Spain & Isabella, queen of Castle
born 15 December 1485 at Alcala de Henares, Madrid
married to Arthur, prince of Wales, on 14 November 1501 at St. Paul’s Cathedral, London
widowed 2 April 1502 at Ludlow Castle, Shropshire
married to Henry VIII, king of England, on 11 June 1509 at Grey Friars Church, Greenwich

Six Children
daughter, born 31 January 1510
Henry (1), duke of Cornwall, born 1 January 1511
Henry (2), duke of Cornwall, born November 1513
son, born December 1514
Mary I, queen of England, born 18 February 1516
daughter, born 10 November 1518

marriage annulled 1533
died 7 January 1536 at Kimbolton Castle
buried at Peterborough Cathedral

Anne Boleyn, marquess of Pembroke
born about 1501/2 or 1507 at Blickling Hall, Norfolk
titled marquess of Pembroke on 1 September 1532
married to Henry VIII, king of England, on 25 January 1533 at Westminster Abbey, London

Three Children
Elizabeth I, queen of England, born 7 September 1533
Henry, duke of Cornwall, born 1534
son, born 29 January 1536

marriage annulled 1536
executed 19 May 1536 at Tower Green, Tower of London
buried at the Chapel of St. Peter ad Vincula, Tower of London

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Shelley Stewart, Jr.

Corporate executive Shelley Stewart, Jr. was born on April 16, 1953 in Glen Cove, New York to Shelley Stewart, Sr. and Verna Stewart. He received his B.S. and M.S. degrees in criminal justice from Northeastern University in Boston, Massachusetts in 1975 and 1978, respectively, and his M.B.A. degree in business administration from the University of New Haven in West Haven, Connecticut in 1990.

In 1982, Stewart joined United Technologies Corporation (UTC) and served as director of worldwide sourcing and in a variety of leadership and operations positions at United’s subsidiaries, Norden Systems, Inc. and Hamilton Sunstrand. Stewart remained with United Technologies Corporation until 2000, when he became vice president of supply chains at the Raytheon Company. He joined Invensys in 2002 as the company’s senior vice president of supply chain, headquartered in London, England. Then, in 2003, Stewart became the chief procurement officer and senior vice president of operational excellence at Tyco Industries. He also served as vice president of Tyco’s supply chain management from 2003 until 2006. Stewart remained in leadership positions at Tyco Industries until 2012, when he joined E.I. du Pont de Nemours and Company (DuPont) as the chief procurement officer and vice president of sourcing and logistics. In 2016, he assumed the responsibility for facilities services and real estate Stewart co-authored Straight To The Bottom Line: An Executive’s Roadmap to World Class Supply Management, which was published in 2005.

In addition to his business career, Stewart has served on the board of directors of Cleco Corporation, Kontoor Brands, and Otis Worldwide, on the board of Governors for the University of New Haven, as board chair of the Howard University’s School of Business, as vice chair of the National Minority Supplier Development Council, and as board president of The Boys and Girls Club of Trenton/Mercer County. He has served on the board of trustees at Howard University and Northeastern University Corporation. Stewart was also appointed to the U.S. Department of Commerce’s National Advisory Council on Minority Business Enterprise, and was a member of The Conference Board's Purchasing and Supply Leadership Council.

Stewart and his wife, Ann C. Stewart, have two children, Sydney and Shelley Stewart, III.

Shelley Stewart, Jr. was interviewed by The HistoryMakers on August 17, 2017.


CBSE Previous Year Question Papers Class 12 Physics 2011 Outside Delhi

CBSE Previous Year Question Papers Class 12 Physics 2011 Outside Delhi Set-I

Question 1.
Define electric dipole moment. Write its S.I. unit. [1]
Answer :
Electric dipole moment : Dipole moment is a measure of strength of electric dipole. It is vector quantity whose magnitude is equal to product of magnitude of charge and the distance between them.
p = q x 2d
SI unit of dipole moment is coulomb-metre (Cm).

Question 2.
Where on the surface of Earth is the angle of dip 90° ? [1]
Answer :
Magnetic dip is the angle made by a compass needle with the horizontal point on earth’s surface. The angle of dip is 90° at the poles.

Question 3.
A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. What is potential at the center of the sphere ? [1]
Answer :
Potential inside the charged sphere is constant and equal to potential on the surface of the conductor. So, potential at the center of sphere is 10 V.

Question 4.
How are radio waves produced ? [1]
Answer:
Radio waves are produced by :

  • Rapid acceleration and deceleration of electrons.
  • Using tuning circuits like LCR, LC and RC.
  • Accelerated motion of charges in conducting wires.
  • They are also given off by stars, sparks and lightning.

Question 5.
Write any two characteristic properties of nuclear force. [1]
Answer :
Characteristic properties of nuclear forces are:

Question 6.
Two bar magnets are quickly moved towards a metallic loop connected across a capacitor ‘C’ as shown in the figure. Predict the polarity of the capacitor. [1]

Answer:
In this situation, a will become positive with respect to b, as current induced is in clockwise direction.

Question 7.
What happens to the width of depletion layer of a p-n junction when it is
(i) forward biased,
(ii) reverse biased ? [1]
Answer :
(i) Forward biased : Potential drop across the junction decreases and diffusion of holes and electrons across the junction increases. It makes the width of the depletion layer smaller.
(ii) Reverse biased : Potential drop across the junction increases and diffusion of holes and electrons across the junction decreases. It makes the width of the depletion layer larger.

Question 8.
Define the term ‘stopping potential’ in relation to photoelectric effect. [1]
Answer :
Stopping potential is the minimum negative (retarding) potential of anode for which photo current stops or becomes zero. It is denoted by Vs. The value of stopping potential is different for different metals but it is independent of the intensity of incident light and depends on the frequency of the incident light.

Question 9.
A thin straight infinitely long conducting wire having charge density λ is enclosed by a cylindrical surface of radius r and length ‘l’ its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder. [2]
Answer :
Charge enclosed by the cylindrical surface q = λl
(egin mathrm, & phi=frac> phi &=frac> end)

Question 10.
Plot a graph showing the variation of Coulomb force (F) versus (left(frac<1>> ight)), where r is the distance between the two charges of each pair of charges : (1 μC, 2 μC) and (2 μC, -3 μC). Interpret the graphs obtained. [2]
Answer :
The following graph shows the variation of Coulomb force (F) versus r.

Question 11.
Write the expression for Lorentz magnetic force on a particle of charge ‘q’ moving with velocity (vec) in a magnetic field (vec) . Show that no work is done by this force on the charged particle. [2]
Answer:
Magnetic Lorentz force is given by

Question 12.
What are eddy currents ? Write any two applications of eddy currents. [2]
Answer :
When a bulk piece of conductor is subjected to changing magnetic flux, the induced current developed in it is called eddy current.
Applications of eddy currents :

  • Magnetic brakes in trains.
  • Electromagnetic damping.
  • Induction furnaces.
  • Electric power meter.

Question 13.
What is sky wave communication ? Why is this mode of propagation restricted to the frequencies only up to few MHz ? [2]

Question 14.
In the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine the potential at point B. [2]

Question 15.
A parallel plate capacitor is being charged by a time varying current. Explain briefly how Ampere’s circuital law is generalized to incorporate the effect due to the displacement current. [2]
Answer:
Gauss’ law states that the electric flux ΦE of a parallel plate capacitor having an area A, and a total charge Q is given by

Question 16.
Net capacitance of three identical capacitors in series is 1 μF. What will be their net capacitance if connected in parallel ? Find the ratio of energy stored in the two configurations if they are both connected to the same source. [2]
Answer:
Net capacitance in series = 1 μF
If C1 = C2 = C3 = C
Let C be the capacitance of each of three capacitors and Cs and Cp be the capacitance of series and parallel combination respectively.

Question 17.
Using the curve for the binding energy per nucleon as a function of mass number A, state clearly how the release in energy in the processes of nuclear fission and nuclear fusion can be explained.

Answer:
The above curve shows that:

  • When a heavy nucleus breaks into two medium sized nuclei (in nuclear fission), the BE/nucleon increases resulting in the release of energy.
  • When two small nuclei combine to form a relatively bigger nucleus in nuclear fusion, BE/nucleon increases, resulting in the release of energy.

Question 18.
In the meter bridge experiment, balance point was observed at J with AJ = l. [2]
(i) The values of R and X were doubled and then interchanged. What would be the new position of balance point ?
(ii) If the galvanometer and battery are interchanged at the balance position, how will the balance point get affected ?

(ii) By interchanging galvanometer and battery, there will be no change in the balance point position.

Question 19.
A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in
(i) a medium of refractive index 1.65,
(ii) a medium of refractive index 1.33. [3]
(a) Will it behave as a converging or a diverging lens in the two cases ?
(b) How will its focal length change in the two media ?



(b) (i) As seen from equation (ii), focal length becomes negative and increases in magnitude.
(ii) As seen from equation (ii), focal length remains positive and increases in magnitude

Question 20.
Draw a plot showing the variation of photoelectric current with collector plate potential for two different frequencies, v1 > v2, of incident radiation having the same intensity. In which case will be stopping potential be higher ? Justify your answer. [3] Answer:

Stopping potential is more for the curve corresponding to the frequency v2 (∵ v1> v2). This is due to the fact that with increase in the frequency, the kinetic energy of emitted photo electrons also increases. Therefore, we need more negative potential to stop these electrons.

Question 21.
Write briefly any two factors which demonstrate the need for modulating a signal. Draw a suitable diagram to show amplitude modulation using a sinusoidal signal as a modulating signal. [3]

Question 22.
Use the mirror equation to show that [3]
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
Answer:
Mirror equation is given as,


Question 23.
Draw a labelled diagram of a full wave rectifier circuit. State its working principle. Show the input-output wave forms. [3]
Answer:

Rectification : Rectification means conversion of ac into dc. A p-n diode acts as a rectifier because an ac changes polarity periodically and a p-n diode conducts only when it is forward biased it does not conduct when it is reverse biased.

Working : The ac input voltage across secondary S1 and S2 changes polarity after each half cycle. Suppose during the first cycle of input ac signal, the terminal S1 is positive relative to centre tap and S2 is negative relative to it. Then diode D1 is forward biased and D2 is reverse biased. Therefore, diode D1 conducts while D2 does not. Thus, the current in load resistance RL is in the same direction for both half cycles of input ac signal and the output current is a continuous series of unidirectional pulses.
In a full wave rectifier, if input frequency is f Hertz, then output frequency will be 2f Hertz because for each cycle of input, two positive half cycles of output are obtained.

Question 24.
(a) Using de-Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second postulate of quantization of energy levels in a hydrogen atom.
(b) The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state ? [3]
Answer :
(a) The second postulate of Bohr atom model says that angular momentum of electron orbiting around the nucleus is quantized, i.e.,
(m v r=frac<2 pi>, ext < where >n=1,2,3, ldots . .)
According to de-Broglie, a stationary orbit is that which contains an integral number of de-Broglie waves associated with the revolving electron.


Question 25.
You are given a circuit below. Write its truth table. Hence, identify the logic operation carried out by this circuit. Draw the logic symbol of the gate it corresponds to. [3]

Question 26.
A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope. [3]
OR
A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1.0 cm is used, find the angular magnification of the telescope.

If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens ? The diameter of the moon is 3.42 x 10 6 m and the radius of the lunar orbit is 3.8 x 10 8 m.




Question 27.
Two heating elements of resistances R1 and R2 when operated at a constant supply of voltage V, consume powers P1 and P2 respectively. Deduce the expressions for the power of their combination when they are, in turn, connected in
(i) series and
(ii) parallel across the same voltage supply. [3]
Answer:
(i) In series,


Question 28.
(a) Using Ampere’s circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis.
(b) In what respect is a toroid different from a solenoid ? Draw and compare the pattern of the magnetic field lines in the two cases.
(c) How is the magnetic field inside a given solenoid made strong ? [5]
OR
(a) State the principle of the working of a moving coil galvanometer, giving its labelled diagram.
(b) “Increasing the current sensitivity of a galvonometer may not necessarily increase its voltage sensitivity.” Justify this statement.
(c) Outline the necessary steps to convert a galvanometer of resistance RG into an ammeter of a given range.
Answer:
(a) Magnetic field inside a long solenoid is uniform every where and approximately zero outside it. Fig. shows a sectional view of long solenoid current coming out of the plane of the papers at points marked (•) and current entering the plane of the paper at points marked (x). To find the magnetic field B at any point inside the solenoid, consider a rectangular loop abed as Amperian loop. According to Ampere’s circuital law,


(b) Difference: In a toroid, magnetic lines do not exist outside the body. Toroid is closed whereas the solenoid is open on both sides. Magnetic field is uniform inside a toroid whereas for solenoid, it is different at the two ends and center.

(c) Strength of magnetic field :
(1) By inserting the ferromagnetic substance inside the solenoid.
(2) By increasing the current through the solenoid.
OR

Principle : Its working is based on the fact that when a current carrying coil is placed in a magnetic field, it experiences a torque.
Working : When current (I) is passed in the coil, torque t acts on the coil, given by
T = NIAB sin θ
where, θ = Angle between normal to plane of coil
B = Magnetic field of strength
N = No. of turns in a coil.
For equilibrium,
deflecting torque = restoring torque
NIAB = Cθ
(Rightarrow quad heta=frac>> mathrm)
where, C = Torsional rigidity of the wire
⇒ θ ∝ I
The deflection of coil is directly proportional to the current flowing in the coil.
(b) Due to deflecting torque, the coil rotates and suspension wire gets twisted. A restoring torque is set up in the suspension wire.


It means voltage sensitivity is dependent on current sensitivity and resistance of galvanometer, R. If we increase current sensitivity and galvanometer resistance is high, then it is not certain that voltage sensitivity will be increased. Thus, the increase of current sensitivity does not imply the increase of voltage sensitivity.
(c) Conversion of a galvanometer to ammeter :

A galvanometer can be converted into ammeter by connecting a shunt (low resistance) in parallel with the galvanometer and its value is given by

Question 29.
State the working of a.c. generator with the help of a labelled diagram. The coil of an a.c. generator having N turns, each of area A, is rotated with a constant angular velocity ω. Deduce the expression for the alternating e.m.f. generated in the coil. What is the source of energy generation in this device ? [5]
OR
(a) Show that in an a.c. circuit containing a pure inductor, the voltage is ahead of current by π/2 in phase.
(b) A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth’s magnetic field B.
(i) Write the expression for the instantaneous value of the e.m.f. induced in the wire.
(ii) What is the direction of the e.m.f. ?
(iii) Which end of the wire is at the higher potential ?
Answer:

Working : When a coil (armature) rotates inside a uniform magnetic field, magnetic flux linked with the coil changes w.r.t. time. This produces an e.m.f. according to Faraday’s law.

For first half of the rotation, the current will be from one end (first ring) to the other end (second ring). For second half of the rotation, it is in opposite sense.

To calculate the magnitude of e.m.f. induced, Suppose
N = number of turns in the coil.
A = area enclosed by each turn of coil.
(vec=) = strength of magnetic field.
θ= angle which normal to the coil makes with (vec=) at any instant t,

Put in (ii), if we denote NABω as e0, then e = e0 sin ωt
Source of energy: Mechanical energy.
The word generator is a misnomer, because nothing is generated by the machine, it is an alternator converting one form of energy to another.
OR
(a) Circuit containing inductance only : Let an alternating emf given by E = E0 sin ωt, …(i)

be applied across a pure (zero resistance) coil of inductance L. As the current i in the coil grow continuously, an opposing emf is induced in the coil whose magnitude is (mathrm frac) where (frac) is the rate of change of current. But this should be zero because there is no resistance in the current. Thus,

From equations (i) and (ii), it is proved that voltage is ahead of current by π/2.

(b) (i) e = BLv
(ii) Direction of e.m.f is from west to east.
(iii) Wire 1 is at greater potential than wire 2.

Question 30.
State the importance of coherent sources in the phenomenon of interference. In Young’s double slit experiment to produce interference pattern, obtain the conditions for constructive and destructive interference. Hence deduce the expression for the fringe width. How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water ? [5]
Answer :
Two sources of light which continuously emit light waves of same frequency with a zero or constant phase difference between them are called coherent sources. They are necessary to produce sustained interference pattern.

A thin film of oil spread over water shows beautiful colors due to interference of light. If coherent sources are not taken, the phase difference between the two interfering waves will change continuously and a sustained interference pattern will not be obtained.


Constructive interference: The intensity of light will be maximum at those places where the path difference between the interfering light waves is zero or an integral multiple of λ, i.e., λ , 2λ ,……….
Hence for maximum intensity, we have

Destructive interference : The intensity of light will be minimum at those places where the path difference between the interfering light-waves is

destructive interference.
So fringe width is directly proportional to λ On immersing the apparatus in water, the wavelength of light decreases (λω = λ/n).
Therefore, fringe width will decrease in water.

CBSE Previous Year Question Papers Class 12 Physics 2011 Outside Delhi Set-II

Note : Except for the following questions, all the remaining questions have been asked in previous Set.

Question 1.
A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 5 V. What is the potential at the centre of the sphere ? [1]
Answer :
Potential inside the charged sphere is constant and equal to potential on the surface of conductor. Therefore, potential at the center of sphere is 5 V.

Question 2.
How are X-rays produced ? 11]
Answer :
X-rays are produced when electron strike a metal target. The electrons are liberated from the heated filament and accelerated from the high voltage towards the metal target. X-rays are also produced when electrons collide with the atom and nuclei of metal target.

Question 3.
Where on the surface of Earth is the angle of dip zero ? [1]
Answer :
At magnetic equator angle of dip is zero.

Question 4.
State the principle of working of a transformer. Can a transformer be used to step up or step down a d.c. voltage ? Justify your answer. [2]
Answer :
Transformer principle : It is a device which converts high voltage a.c. into low voltage a.c. and vice-versa.

It is based upon the principle of mutual induction. When alternating current is passed through a coil, an induced e.m.f. is set up in the neighboring coil.

Working: When an alternating current is passed through the primary, the magnetic flux through the iron core changes which does two things. It produces e.m.f. in the primary and an induced e.m.f. is also set up in the secondary. If we assume that the resistance of primary is negligible, the back e.m.f. will be equal to the voltage applied to the primary.

A transformer can not be used to step up or step down a d.c. voltage because d.c. can not produce a changing magnetic flux in the core of the transformer and no emf will be induced.

Question 5.
In the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine the potential at point B. [2]

Question 6.
What is ground wave communication ? On what factors does the maximum range of propagation in this mode depend ? [2]

Question 7.
A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in
(i) a medium of refractive index 1.6,
(ii) a medium of refractive index 1.3. [3]
(a) Will it behave as a converging or a diverging lens in two cases ?
(b) How will its focal length change in the two media ?



CBSE Previous Year Question Papers Class 12 Physics 2011 Outside Delhi Set-III

Note : Except for the following questions, all the remaining questions have been asked in previous Set.

Question 1.
A hollow metal sphere of radius 6 cm is charged such that the potential on its surface is 12 V. What is the potential at the centre of sphere ? [1]
Answer :
Potential inside the charged sphere is constant and equal to potential on the surface of conductor. So therefore, potential at the centre of sphere is 12 V.

Question 2.
How are microwaves produced ? [1]
Answer :
Microwaves are electromagnetic waves with wavelength ranging from as long as meter to as short as one millimeter, or equivalently with frequencies between 300 MFz and 300 GHz. Microwaves are produced by vacuum tubes devices that operate on the ballistic motion of electron controlled by magnetic or electric fields. Some different kinds of microwaves emitters are the cavity magnetron, the klystron, the travelling wave tube (TWT), the gyrotron and all stars.

Question 3.
Mention various energy losses in transformer. [2]
Answer :
Magnetic core losses are exaggerated with higher frequencies, eddy currents in the iron core, resistance of windings or copper loss, hysteresis loss and flux leakage are energy losses in transformers. Transformers energy losses tend to worsen with increasing frequency.

Question 4.
In the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine the potential at point B. [2]

Question 5.
A thin straight infinitely long conducting wire having charge density λ is enclosed by a cylindrical surface of radius r and length l, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder. [2]
Answer:

There will be no electric flux through the circular ends of the cylinder.
So, according to Gauss’s law,
(mathrm, quad phi=frac>)
Since, charge enclosed by gaussian surface
(egin ext < i.e. >& q=lambda imes l herefore & phi=frac> end)

Question 6.
A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. It is immersed in a liquid of refractive index 1.3. Calculate its new focal length. [3]


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