100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	During Medieval Period, which of the following language was designated as the 'Camp Language'?1. Sanskrit2. Urdu3. Hindi4. Persian
Answer» Correct Answer - Option 2 : Urdu The correct answer is Urdu. Urdu was designated the Camp language of Medieval India The word, Turkish word Urdu means military camp 'a Lashkar' The Urdu language was developed due to contact with Hindu, Turkish and Afghan soldiers.

1.

During Medieval Period, which of the following language was designated as the 'Camp Language'?1. Sanskrit2. Urdu3. Hindi4. Persian

Answer» Correct Answer - Option 2 : Urdu

The correct answer is Urdu.

Urdu was designated the Camp language of Medieval India
The word, Turkish word Urdu means military camp 'a Lashkar'
The Urdu language was developed due to contact with Hindu, Turkish and Afghan soldiers.

Discussion

2.	Akbar's Din-i-illahi was called as a monument of his folly, not of wisdom by which writer'?1. Badayuni2. Vinset Smith3. Barni4. ibn battuta
Answer» Correct Answer - Option 2 : Vinset Smith The correct answer is Vinset Smith. Akbar was born as Abu'l-Fath Jalal ud-din Muhammad Akbar on October 15, 1542. Akbar succeeded his father, Humayun, in 1556 when he was 13 years old. Akbar was given the nickname 'the Great' because of his many accomplishments. Akbar started a new religion that was based on the teachings of major religions like Hinduism, Christianity, Zoroastrianism, etc. This religious path was later referred to as the Din-i-illahi or Divine Monotheism. Historian Vinset Smith called Akbar's Din-i-ilahi as a monument of his folly not of wisdom.

2.

Akbar's Din-i-illahi was called as a monument of his folly, not of wisdom by which writer'?1. Badayuni2. Vinset Smith3. Barni4. ibn battuta

Answer» Correct Answer - Option 2 : Vinset Smith

The correct answer is Vinset Smith.

Akbar was born as Abu'l-Fath Jalal ud-din Muhammad Akbar on October 15, 1542.
Akbar succeeded his father, Humayun, in 1556 when he was 13 years old.
Akbar was given the nickname 'the Great' because of his many accomplishments.
Akbar started a new religion that was based on the teachings of major religions like Hinduism, Christianity, Zoroastrianism, etc.
This religious path was later referred to as the Din-i-illahi or Divine Monotheism.
Historian Vinset Smith called Akbar's Din-i-ilahi as a monument of his folly not of wisdom.

Discussion

3.	0. गॉउस का नियम लिखिए। किसी एक समान आवेशित पतले गोलीय कोश के कारण किसी बिंदु पर विद्युत क्षेत्र की तीव्रता की गणना कीजिए जबकि -1. बिंदु गोलीय कोश के अंदर है।2. बिंदु गोलीय कोष के बाहर हैआवश्यक चित्र बताइए।
Answer» गाउस के नियम: “किसी विद्युत क्षेत्र में उपस्थित काल्पनिक या स्वेच्छा गृहीत बन्द पृष्ठ से अभिलंबवत बाहर निकलने वाला कुल विद्युत फ्लक्स उस बंद पृष्ठ द्वारा परिबद्ध कुल आवेश का 1/ε₀K गुना होता है।

3.

0. गॉउस का नियम लिखिए। किसी एक समान आवेशित पतले गोलीय कोश के कारण किसी बिंदु पर विद्युत क्षेत्र की तीव्रता की गणना कीजिए जबकि -1. बिंदु गोलीय कोश के अंदर है।2. बिंदु गोलीय कोष के बाहर हैआवश्यक चित्र बताइए।

Answer»

गाउस के नियम:

“किसी विद्युत क्षेत्र में उपस्थित काल्पनिक या स्वेच्छा गृहीत बन्द पृष्ठ से अभिलंबवत बाहर निकलने वाला कुल विद्युत फ्लक्स उस बंद पृष्ठ द्वारा परिबद्ध कुल आवेश का 1/ε₀K गुना होता है।

Discussion

4.	The principal value of sin-1(sin 2π/3) is(A) -2π/3(B) 2π/3(C) 4π/3(D) π/3
Answer» correct option: (D) π/3

4.

The principal value of sin-1(sin 2π/3) is(A) -2π/3(B) 2π/3(C) 4π/3(D) π/3

Answer»

correct option:

(D) π/3

Discussion

5.	Find the roots of the quadratic equation 2x2–x – 6 = 0.
Answer» Given quadratic equation is : 2x² – x – 6 = 0. ∴ a = 2, b = – 1 & c = – 6. ∴ Roots of given quadratic equation are x = \(\frac{-b±\sqrt{b^2-4ac}}{{2a}}\) = \(\frac{-(-1)±\sqrt{(-1)^2-4\times 2\times -6}}{{2\times 2}}\) = \(\frac{1±\sqrt{1+48}}{{4}}\) = \(\frac{1±7}{4}\) = \(\frac{1+7}{4}\) and \(\frac{1-7}{4}\) = 2 and \(\frac{-6}{4}\) = 2 and \(\frac{-3}{2}\). Hence, The roots of the given equations are 2 and \(\frac{-3}{2}\).

5.

Find the roots of the quadratic equation 2x2–x – 6 = 0.

Answer»

Given quadratic equation is :

2x² – x – 6 = 0.

∴ a = 2, b = – 1 & c = – 6.

∴ Roots of given quadratic equation are x = \(\frac{-b±\sqrt{b^2-4ac}}{{2a}}\)

= \(\frac{-(-1)±\sqrt{(-1)^2-4\times 2\times -6}}{{2\times 2}}\)

= \(\frac{1±\sqrt{1+48}}{{4}}\)

= \(\frac{1±7}{4}\)

= \(\frac{1+7}{4}\) and \(\frac{1-7}{4}\)

= 2 and \(\frac{-6}{4}\)

= 2 and \(\frac{-3}{2}\).

Hence,

The roots of the given equations are 2 and \(\frac{-3}{2}\).

Discussion

6.	Examine if the set is subspace of R3 s = {(x, y, z) € R3 : x= 0}
Answer» Let \(V_1, V_2 \in S\) such that \(V_1 = (x_1, y_1, z_1) \) & \(V_2 = (x_2, y_2, z_2) \) then x₁ = 0 and x₂ = 0. Let \(a \in R\) Now, \(aV_1 + V_2 = ax_1,y_1, z_1) + (x_2,y_2,z_2)\) \(= (ax_1 + x_2, ay_1+y_2, az_1+ z_2)\in S\) \((\because ax_1 + x_2 = a\times 0 + 0 = 0)\) \(\therefore\) S is subspace of \(R^3\).

6.

Examine if the set is subspace of R3 s = {(x, y, z) € R3 : x= 0}

Answer»

Let \(V_1, V_2 \in S\) such that

\(V_1 = (x_1, y_1, z_1) \) & \(V_2 = (x_2, y_2, z_2) \)

then x₁ = 0 and x₂ = 0.

Let \(a \in R\)

Now,

\(aV_1 + V_2 = ax_1,y_1, z_1) + (x_2,y_2,z_2)\)

\(= (ax_1 + x_2, ay_1+y_2, az_1+ z_2)\in S\)

\((\because ax_1 + x_2 = a\times 0 + 0 = 0)\)

\(\therefore\) S is subspace of \(R^3\).

Discussion

7.	The product of three numbers in G.P. is 216. If 2, 8, 6 be added to them, the results are in A.P. Find the numbers.
Answer» Let the three numbers be \(\frac{a}{r}\),a, ar. ∴ According to the question ⇒ \(\frac{a}{r} \times a \times ar\) = 216 .......(1) ⇒ a³ = 216 ⇒ a = 6 2,8,6 is added to them \(\therefore\) \(\frac{a}{r} + 2\), a + 8, ar + 6 The above sequence is in AP. We know in AP. 2b = a + c ⇒ 2(a + 8) = \(\frac{a}{r}\) + 2 + ar + 6 Substituting a = 6 in above equation we get, ⇒ 2(6 + 8) = \(\frac{6}{r}\) + 2 + 6r + 6 ⇒ 28 = \(\frac{6 + 6r^2 + 8r}{r}\) ⇒ 28r = 6 + 6r² + 8r ⇒ 6r² - 18r - 2r - 6 = 0 ⇒ 6r(r-3) - 2(r - 3) = 0 ⇒ r = 3 or r = 1/3 ∴ Now the equation will be ⇒ \(\frac{6}{3}, 6,6 \times 3\) or \(\frac{6}{\frac{1}{2}}, 6 ,6 \times \frac{1}{3}\) ⇒ 2,6,18 or 18,6,2 ∴ The three numbers are 2,6,18.

7.

The product of three numbers in G.P. is 216. If 2, 8, 6 be added to them, the results are in A.P. Find the numbers.

Answer»

Let the three numbers be \(\frac{a}{r}\),a, ar.

∴ According to the question

⇒ \(\frac{a}{r} \times a \times ar\) = 216 .......(1)

⇒ a³ = 216

⇒ a = 6

2,8,6 is added to them

\(\therefore\) \(\frac{a}{r} + 2\), a + 8, ar + 6

The above sequence is in AP.

We know in AP.

2b = a + c

⇒ 2(a + 8) = \(\frac{a}{r}\) + 2 + ar + 6

Substituting a = 6 in above equation we get,

⇒ 2(6 + 8) = \(\frac{6}{r}\) + 2 + 6r + 6

⇒ 28 = \(\frac{6 + 6r^2 + 8r}{r}\)

⇒ 28r = 6 + 6r² + 8r

⇒ 6r² - 18r - 2r - 6 = 0

⇒ 6r(r-3) - 2(r - 3) = 0

⇒ r = 3 or r = 1/3

∴ Now the equation will be

⇒ \(\frac{6}{3}, 6,6 \times 3\) or \(\frac{6}{\frac{1}{2}}, 6 ,6 \times \frac{1}{3}\)

⇒ 2,6,18 or 18,6,2

∴ The three numbers are 2,6,18.

Discussion

8.	A ball is dropped from a height of `20 m` and rebounds with a velocity which is `3/4` of the velocity with which it hits the ground. What is the time interval between the first and second bunces `(g=10m//s^(2))`A. `3` secB. `4` secC. `5` secD. `6` sec
Answer» Correct Answer - A Velocity after collision `v=(3)/(4) xx sqrt(2xx10 xx 20)=15 m//s t=(2u)/(g)=3sec`.

Discussion

9.	X sent out goods costing Rs.80,000 to Y so as to show 20% profit on invoice price. 40% of the goods were lost in transit. 60% of the goods received by consignee was sold at 25% above invoice price. Rate of commission is 10% on sales at invoice price plus 50% of the surplus over invoice price. Calculate the commission due to Y.
Answer» Invoice Price of Goods Sold = 60% (Rs.1,00,000 – 40% of Rs.1,00,000) = Rs.36,000 Surplus over invoice price = Rs. 36,000 × 25% = Rs.9000 Commission = (36,000 × 10/100) + (9,000 × 50/100) = Rs. 8,100.

9.

X sent out goods costing Rs.80,000 to Y so as to show 20% profit on invoice price. 40% of the goods were lost in transit. 60% of the goods received by consignee was sold at 25% above invoice price. Rate of commission is 10% on sales at invoice price plus 50% of the surplus over invoice price. Calculate the commission due to Y.

Answer»

Invoice Price of Goods Sold = 60% (Rs.1,00,000 – 40% of Rs.1,00,000) = Rs.36,000

Surplus over invoice price = Rs. 36,000 × 25% = Rs.9000

Commission = (36,000 × 10/100) + (9,000 × 50/100) = Rs. 8,100.

Discussion

10.	If Raj bought an umbrella for Rs. 100 and sold it for a profit of Rs. 10 then find the selling price of umbrella?1. Rs. 1152. Rs. 1003. Rs. 904. Rs. 110
Answer» Correct Answer - Option 4 : Rs. 110 Given: C.P of umbrella = Rs. 100 Profit = Rs. 10 Formula used: S.P = C.P + Profit Calculation: Selling price = Rs. 100 + Rs. 10 ⇒ Rs. 110 ∴ The selling price of umbrella is Rs. 110

10.

If Raj bought an umbrella for Rs. 100 and sold it for a profit of Rs. 10 then find the selling price of umbrella?1. Rs. 1152. Rs. 1003. Rs. 904. Rs. 110

Answer» Correct Answer - Option 4 : Rs. 110

Given:

C.P of umbrella = Rs. 100

Profit = Rs. 10

Formula used:

S.P = C.P + Profit

Calculation:

Selling price = Rs. 100 + Rs. 10

⇒ Rs. 110

∴ The selling price of umbrella is Rs. 110

Discussion

11.	After a discount of 37.5% an article is sold for Rs. 4545. What is the marked price of the article?1. Rs. 45582. Rs. 72723. Rs. 50004. Rs. 6500
Answer» Correct Answer - Option 2 : Rs. 7272 Given: After a discount of 37.5% an article is sold for Rs.4545. Calculation: 37.5% = 3/8 in fraction form According to the question, MP × 5/8 = 4545 where MP = marked price ⇒ MP = (4545 × 8)/5 ⇒ MP = 909 × 8 = 7272 ∴ The marked price of an article is Rs.7272.

11.

After a discount of 37.5% an article is sold for Rs. 4545. What is the marked price of the article?1. Rs. 45582. Rs. 72723. Rs. 50004. Rs. 6500

Answer» Correct Answer - Option 2 : Rs. 7272

Given:

After a discount of 37.5% an article is sold for Rs.4545.

Calculation:

37.5% = 3/8 in fraction form

According to the question,

MP × 5/8 = 4545

where MP = marked price

⇒ MP = (4545 × 8)/5

⇒ MP = 909 × 8 = 7272

∴ The marked price of an article is Rs.7272.

Discussion

12.	A man purchased 150 t-shirts, each costing the same, but 40% of them are lost in transit which cannot be recovered. He sold 50% of the remaining at 20% profit each and remaining at 5% loss each. If the total selling price of t-shirts is Rs. 7740, then what was the cost price of each t-shirts?1. Rs. 1002. Rs. 603. Rs. 904. Rs. 805. Rs. 85
Answer» Correct Answer - Option 4 : Rs. 80 Given: Total number of t-shirts = 150 Lost in transit = 40% Formula used: SP = [CP + (CP × profit%)] SP = [CP – (CP × loss%)] SP = selling price CP = cost price Calculation: Let CP be Rs. 20x According to the question, 40% of the t-shirts are lost in transit So, only 150 × 60% = 90 t-shirts can be sold 45 t-shirts at 20% profit ⇒ 45 × 20x × 6/5 ⇒ 45 × 24x ⇒ 1080x 45 t-shirt at 5% loss ⇒ 45 × 20x × 19/20 ⇒ 45 × 19x ⇒ 855x 1080x + 855x = 7740 ⇒ 1935x = 7740 ⇒ x = 4 ⇒ 20x = 80 ⇒ CP = Rs. 80 ∴ Cost price of each t-shirts is Rs. 80

12.

A man purchased 150 t-shirts, each costing the same, but 40% of them are lost in transit which cannot be recovered. He sold 50% of the remaining at 20% profit each and remaining at 5% loss each. If the total selling price of t-shirts is Rs. 7740, then what was the cost price of each t-shirts?1. Rs. 1002. Rs. 603. Rs. 904. Rs. 805. Rs. 85

Answer» Correct Answer - Option 4 : Rs. 80

Given:

Total number of t-shirts = 150

Lost in transit = 40%

Formula used:

SP = [CP + (CP × profit%)]

SP = [CP – (CP × loss%)]

SP = selling price

CP = cost price

Calculation:

Let CP be Rs. 20x

According to the question,

40% of the t-shirts are lost in transit

So, only 150 × 60% = 90 t-shirts can be sold

45 t-shirts at 20% profit

⇒ 45 × 20x × 6/5

⇒ 45 × 24x

⇒ 1080x

45 t-shirt at 5% loss

⇒ 45 × 20x × 19/20

⇒ 45 × 19x

⇒ 855x

1080x + 855x = 7740

⇒ 1935x = 7740

⇒ x = 4

⇒ 20x = 80

⇒ CP = Rs. 80

∴ Cost price of each t-shirts is Rs. 80

Discussion

13.	Marked price of Bajaj iron, Usha iron and Philips iron are Rs.2400, Rs.3000 and Rs.3600 respectively. Bajaj iron is sold at a discount of 5%, Usha iron at a discount of 8% and Philips iron at a discount of 10%, yet the shopkeeper got an overall profit of 20%. If cost price of Bajaj iron is half of Usha iron and the cost price of Usha iron is half of Philips iron, find the cost price of Usha iron.1. Rs. 19712. Rs. 19733. Rs. 19704. Rs. 19755. Rs. 1981
Answer» Correct Answer - Option 1 : Rs. 1971 Given: SP of Bajaj iron = Rs. 2400 SP of Usha iron = Rs. 3000 SP of Philips iron = Rs. 3600 Concept used: SP = [CP – (CP × Discount%)] Calculation: Let CP of Bajaj iron be 100x Then, CP of Usha iron = 200x CP of Philips iron = 400x SP of Bajaj iron = Rs. 2400 × 19/20 ⇒ Rs. 2280 SP of Usha iron = Rs. 3000 × 23/25 ⇒ Rs. 2760 SP of Philips iron = Rs. 3600 × 9/10 ⇒ Rs. 3240 According to the question, (100x + 200x + 400x) × 6/5 = (2280 + 2760 + 3240) ⇒ 700x × 6/5 = 8280 ⇒ 840x = 8280 ⇒ x = 8280/840 ⇒ 200x = [(8280/840) × 200] ⇒ 1971.42 ≈ 1971 ⇒ CP of Usha iron = Rs. 1971 ∴ Cost price of Usha iron is Rs. 1971

13.

Marked price of Bajaj iron, Usha iron and Philips iron are Rs.2400, Rs.3000 and Rs.3600 respectively. Bajaj iron is sold at a discount of 5%, Usha iron at a discount of 8% and Philips iron at a discount of 10%, yet the shopkeeper got an overall profit of 20%. If cost price of Bajaj iron is half of Usha iron and the cost price of Usha iron is half of Philips iron, find the cost price of Usha iron.1. Rs. 19712. Rs. 19733. Rs. 19704. Rs. 19755. Rs. 1981

Answer» Correct Answer - Option 1 : Rs. 1971

Given:

SP of Bajaj iron = Rs. 2400

SP of Usha iron = Rs. 3000

SP of Philips iron = Rs. 3600

Concept used:

SP = [CP – (CP × Discount%)]

Calculation:

Let CP of Bajaj iron be 100x

Then, CP of Usha iron = 200x

CP of Philips iron = 400x

SP of Bajaj iron = Rs. 2400 × 19/20

⇒ Rs. 2280

SP of Usha iron = Rs. 3000 × 23/25

⇒ Rs. 2760

SP of Philips iron = Rs. 3600 × 9/10

⇒ Rs. 3240

According to the question,

(100x + 200x + 400x) × 6/5 = (2280 + 2760 + 3240)

⇒ 700x × 6/5 = 8280

⇒ 840x = 8280

⇒ x = 8280/840

⇒ 200x = [(8280/840) × 200]

⇒ 1971.42 ≈ 1971

⇒ CP of Usha iron = Rs. 1971

∴ Cost price of Usha iron is Rs. 1971

Discussion

14.	A person sold an article from ₹ 3600 and got a profit of20%. Had he sold the article for ₹ 3150, how much profit would he have got?A. 0.04B. 0.05C. 0.06D. 0.1
Answer» Correct Answer - B Let the cost price of the article be ₹x. After 20% profit `rArr (120x)/(100) = 3600` x = 300 Now, profit precentage, when the article is sold for ₹ 3150 `= (3150-3000)/(3000) xx 100 - (150)/(3000) xx 100 = 5%`

Discussion

15.	A man wanted to sell an article with 20% profit, but he actually sold at 20% loss for Rs. 480, At what price he wanted to sell it to earn the profit?1. Rs.7202. Rs.6003. Rs.8404. Rs.750
Answer» Correct Answer - Option 1 : Rs.720 Given: When he sold at 20% loss, Selling price = Rs. 480 Profit % = 20% Formula used: Profit = S.P – C.P Loss = C.P – S.P Profit % = (profit/C.P) × 100 Loss % = (loss/C.P) × 100 Calculation: Let the cost price be 'x'. Selling Price initially = Rs. 480 Loss = 20% of x ⇒ Loss = x/5 S.P = C.P + loss ⇒ S.P = x – x/5 ⇒ (4x)/5 = 480 ⇒ x = 600 Profit = 20% of 600 ⇒ Profit = Rs. 120 New selling price = C.P + profit ⇒ New selling price = 600 + 120 ⇒ Rs. 720 ∴ The new selling price when 20% profit gain is Rs. 720.

15.

A man wanted to sell an article with 20% profit, but he actually sold at 20% loss for Rs. 480, At what price he wanted to sell it to earn the profit?1. Rs.7202. Rs.6003. Rs.8404. Rs.750

Answer» Correct Answer - Option 1 : Rs.720

Given:

When he sold at 20% loss, Selling price = Rs. 480

Profit % = 20%

Formula used:

Profit = S.P – C.P

Loss = C.P – S.P

Profit % = (profit/C.P) × 100

Loss % = (loss/C.P) × 100

Calculation:

Let the cost price be 'x'.

Selling Price initially = Rs. 480

Loss = 20% of x

⇒ Loss = x/5

S.P = C.P + loss

⇒ S.P = x – x/5

⇒ (4x)/5 = 480

⇒ x = 600

Profit = 20% of 600

⇒ Profit = Rs. 120

New selling price = C.P + profit

⇒ New selling price = 600 + 120

⇒ Rs. 720

∴ The new selling price when 20% profit gain is Rs. 720.

Discussion

16.	Kartik sold an item for ₹ 6,500 and incurred a loss of 20%. At what price should he have sold the item to have gained a profit of20%?A. ₹10,375B. ₹9,750C. ₹8,125D. Cannot be determined
Answer» Correct Answer - B Let Cost Price item be x Its selling Price `= x -(20)/(100) x = 6500 (80)/(100) = 6500` `x = 65000 xx (100)/(80) = 8125` S.P of item to have gained a Profit of 20%. `= 8125+(20)/(100) xx 8125 = 9750`

Discussion

17.	If the ratio of cost price to selling price is 10: 11, then profit percent is:1. 1.1%2. 1%3. 10%4. 0.1%
Answer» Correct Answer - Option 3 : 10% Given The ratio of cost price to selling price is 10: 11 Formula used Profit% = {(CP - SP)/CP} × 100 Calculation ⇒ CP = 10 ⇒ SP = 11 ⇒ Profit = 11 - 10 ⇒ Rs.1 ⇒ Profit% = {1/10} × 100 ⇒ 10%

17.

If the ratio of cost price to selling price is 10: 11, then profit percent is:1. 1.1%2. 1%3. 10%4. 0.1%

Answer» Correct Answer - Option 3 : 10%

Given

The ratio of cost price to selling price is 10: 11

Formula used

Profit% = {(CP - SP)/CP} × 100

Calculation

⇒ CP = 10

⇒ SP = 11

⇒ Profit = 11 - 10

⇒ Rs.1

⇒ Profit% = {1/10} × 100

⇒ 10%

Discussion

18.	Find the words which are out of the logic list:A) canteen B) dining-hall C) warehouse D) kitchen E) buffet
Answer» Correct option is C) warehouse

Discussion

19.	If the Cost price of 25 books is equal to the selling price of 20 book. Find the profit percent.1. 20%2. 25%3. 18%4. 15%5. 8%
Answer» Correct Answer - Option 2 : 25% Given Cost price of 25 books is equal to the selling price of 20 book Formula used Profit = {(SP - CP)/CP} × 100 Calculation CP of 25 books is same as SP of 20 books ⇒ 25CP = 20SP ⇒ CP/SP = 20/25 ⇒ Profit = 25 - 20 = 5 ⇒ Profit% = (5/20) × 100 ∴ Profit is 25%

19.

If the Cost price of 25 books is equal to the selling price of 20 book. Find the profit percent.1. 20%2. 25%3. 18%4. 15%5. 8%

Answer» Correct Answer - Option 2 : 25%

Given

Cost price of 25 books is equal to the selling price of 20 book

Formula used

Profit = {(SP - CP)/CP} × 100

Calculation

CP of 25 books is same as SP of 20 books

⇒ 25CP = 20SP

⇒ CP/SP = 20/25

⇒ Profit = 25 - 20 = 5

⇒ Profit% = (5/20) × 100

∴ Profit is 25%

Discussion

20.	Biological value of soyabean protein is (A) 86 (B) 71 (C) 64 (D) 54
Answer» Biological value of soyabean protein is 64.

Discussion

21.	The chemical score of different proteins is calculated in terms of (A) Egg proteins (B) Milk proteins (C) Fish proteins (D) Wheat proteins
Answer» (A) Egg proteins

Discussion

22.	Biological value of egg protein is (A) 94 (B) 60 (C) 51 (D) 40
Answer» Biological value of egg protein is 94.

Discussion

23.	Hepato-pancreatic duct opens into the duodenum and carries (a) Bile (b) Pancreatic juice (c) Both bile and pancreatic juice (d) Saliva
Answer» (c) Both bile and pancreatic juice

Discussion

24.	Which one of the following is a basic amino acid ?A. `{:(" "COOH),(" "\|),(H_(2)N-C-H),(" "\|),(" "CH(CH_(3))_(2)):}`B. `{:(" "COO H),(" "\|),(H_(2)N-C-H),(" "\|),(" "(CH_(2))_(4)NH_(2)):}`C. `{:(" "COO H),(" "\|),(H_(2)N-C-H),(" "\|),(" "CH_(2)OH):}`D. `{:(" "COO H),(" "\|),(H_(2)N-C-H),(" "\|),(" "CH_(3)):}`
Answer» Correct Answer - B

Discussion

25.	Which of the following is/are the characteristics of the family whose floral diagram is given ? (i) Aloe, Asparagus and Colchicum belong to this family (ii) Flower is zygomorphic (iii) Floral formula is (iv) Fruits are berry or capsuleA. (i),(ii) and (iii)B. (iii) onlyC. (i) onlyD. (i),(iii) and (iv)
Answer» Correct Answer - D

Discussion

26.	Select an incorrect option regarding the given figure A. P is a compound gland with exocrine and endocrine sectionsB. R is formed from duct of gall bladder along with the hepatic ductC. R is guarded by a sphincter of OdiiD. Q stores bile which is actually produced in liver
Answer» Correct Answer - B

Discussion

27.	In cockroaches, digestive juice is secreted by theA. gizzardB. Malpighian tubulesC. hepatic caecaD. oesophagus
Answer» Correct Answer - C

Discussion

28.	लंगर संस्था, जिसे गुरु नानक देव जी ने आरम्भ किया था, गुरु अंगद देव जी ने उसे जारी रखा। गुरु अमरदास जी के समय में भी यह संस्था विस्तृत रूप से जारी रही। उनके लंगर में ब्राह्मण, क्षत्रीय, वैश्य तथा शूद्रों को बिना किसी भेद-भाव के एक ही पंगत में बैठ कर इकट्ठे लंगर छकना पड़ता था। गुरु जी की आज्ञानुसार लंगर किए बगैर कोई भी उन्हें नहीं मिल सकता था। मुगल सम्राट अकबर तथा हरीपुर के राजा को भी गुरु साहिब को मिलने से पहले लंगर में से भोजन छकना पड़ा था। इस तरह से यह संस्था सिक्ख धर्म के प्रचार का एक शक्तिशाली साधन सिद्ध हुई।(a) लंगर प्रथा से क्या भाव है?(b) मंजी-प्रथा से क्या भाव है तथा इसका क्या उद्देश्य था?
Answer» (a) लंगर प्रथा अथवा पंगत से भाव उस प्रथा से है जिसके अनुसार सभी जातियों के लोग बिना किसी भेदभाव के एक ही पंगत में इकट्ठे बैठकर खाना खाते थे। (b) मंजी प्रथा की स्थापना गुरु अमरदास जी ने की थी। उनके समय में सिक्खों की संख्या काफ़ी बढ़ चुकी थी। परन्तु गुरु जी की आयु अधिक होने के कारण उनके लिए एक स्थान से दूसरे स्थान पर जाकर अपनी शिक्षाओं का प्रचार करना कठिन हो गया था। अत: उन्होंने अपने सारे आध्यात्मिक प्रदेशों को 22 भागों में बांट दिया। इनमें से प्रत्येक भाग को ‘मंजी’ कहा जाता था। प्रत्येक मंजी छोटे-छोटे स्थानीय केन्द्रों में बंटी हुई थी जिन्हें पीड़ियां (Piris) कहते थे। मंजी प्रणाली का सिक्ख धर्म के इतिहास में विशेष महत्त्व है। डॉ० गोकुल चन्द नारंग के शब्दों में, “गुरु जी के इस कार्य ने सिक्ख धर्म की नींव सुदृढ़ करने तथा देश के सभी भागों में प्रचार कार्य को बढ़ाने में विशेष योगदान दिया।”

28.

लंगर संस्था, जिसे गुरु नानक देव जी ने आरम्भ किया था, गुरु अंगद देव जी ने उसे जारी रखा। गुरु अमरदास जी के समय में भी यह संस्था विस्तृत रूप से जारी रही। उनके लंगर में ब्राह्मण, क्षत्रीय, वैश्य तथा शूद्रों को बिना किसी भेद-भाव के एक ही पंगत में बैठ कर इकट्ठे लंगर छकना पड़ता था। गुरु जी की आज्ञानुसार लंगर किए बगैर कोई भी उन्हें नहीं मिल सकता था। मुगल सम्राट अकबर तथा हरीपुर के राजा को भी गुरु साहिब को मिलने से पहले लंगर में से भोजन छकना पड़ा था। इस तरह से यह संस्था सिक्ख धर्म के प्रचार का एक शक्तिशाली साधन सिद्ध हुई।(a) लंगर प्रथा से क्या भाव है?(b) मंजी-प्रथा से क्या भाव है तथा इसका क्या उद्देश्य था?

Answer»

(a) लंगर प्रथा अथवा पंगत से भाव उस प्रथा से है जिसके अनुसार सभी जातियों के लोग बिना किसी भेदभाव के एक ही पंगत में इकट्ठे बैठकर खाना खाते थे।

(b) मंजी प्रथा की स्थापना गुरु अमरदास जी ने की थी। उनके समय में सिक्खों की संख्या काफ़ी बढ़ चुकी थी। परन्तु गुरु जी की आयु अधिक होने के कारण उनके लिए एक स्थान से दूसरे स्थान पर जाकर अपनी शिक्षाओं का प्रचार करना कठिन हो गया था। अत: उन्होंने अपने सारे आध्यात्मिक प्रदेशों को 22 भागों में बांट दिया। इनमें से प्रत्येक भाग को ‘मंजी’ कहा जाता था। प्रत्येक मंजी छोटे-छोटे स्थानीय केन्द्रों में बंटी हुई थी जिन्हें पीड़ियां (Piris) कहते थे। मंजी प्रणाली का सिक्ख धर्म के इतिहास में विशेष महत्त्व है। डॉ० गोकुल चन्द नारंग के शब्दों में, “गुरु जी के इस कार्य ने सिक्ख धर्म की नींव सुदृढ़ करने तथा देश के सभी भागों में प्रचार कार्य को बढ़ाने में विशेष योगदान दिया।”

Discussion

29.	State whether the following statements are true or false. Correct the false statements:1. Climate is a long term predominant condition of the atmosphere.2. Climate changes continuously.3. If present climatic conditions are analysed with reference to the past climatic conditions, we can predict climatic changes of the future.4. Forecasting is difficult for places where climatic changes are slow and of a limited nature.5. Climate plays a very important role in the formation and enrichment of soil.
Answer» 1. True. 2. False. Climate does not change continuously. It remains constant in a region for a long duration. 3. True. 4. False. Forecasting is easy for places where climatic changes are slow and of a limited nature. 5. True.

29.

State whether the following statements are true or false. Correct the false statements:1. Climate is a long term predominant condition of the atmosphere.2. Climate changes continuously.3. If present climatic conditions are analysed with reference to the past climatic conditions, we can predict climatic changes of the future.4. Forecasting is difficult for places where climatic changes are slow and of a limited nature.5. Climate plays a very important role in the formation and enrichment of soil.

Answer»

1. True.

2. False. Climate does not change continuously. It remains constant in a region for a long duration.

3. True.

4. False. Forecasting is easy for places where climatic changes are slow and of a limited nature.

5. True.

Discussion

30.	Fractions 3/4÷3
Answer» =3/4÷3 =3/4×1/3 =1/4 ans Q=3/4÷3 =3/4÷3/1 =3/4×1/3 =1/4

30.

Fractions 3/4÷3

Answer»

=3/4÷3

=3/4×1/3

=1/4 ans

Q=3/4÷3

=3/4÷3/1

=3/4×1/3

=1/4

Discussion

31.	If we want to live our life comfortably and think the same for future generations then we should protect our forests properly. Planting of plants is called afforestation. We can plant saplings or barren land or non-agricultural land without converting them into forests. You can improve the deteriorating state of forests by planting 311 social forests, agricultural forests, commercial forests. The government should make every effort to awaken the people regarding the importance of forests and cooperation for their protection. Laws should be strictly enforced by the Forest Department and punish the culprits for cutting down trees, either illegally or without adopting the right method.(A) What are the benefits of forests?(B) How can we save the dwindling forests?
Answer» (A) Forests control the climate compact, forests prevent temperature rise in summer and increase the temperature in winter. The roots of dense forests help in reducing the speed of flowing water, thus reducing the speed of flowing water, thus reducing the outbreak of floods. The roots of trees strengthen the soil tightness and prevent soil erosion. The soil gets green manure in the form of bacteria due to dry leaves falling. (B) We can imporove the deteriorating state of forests continuously by dividing forests into different parts. People should be awakened from time to time for the benefits and care of the forests. Cutting of forests should be declared as a legal offence and strict punishment should be given to those who cut down trees without proper law. People should be constantly encouraged to plant new trees.

31.

If we want to live our life comfortably and think the same for future generations then we should protect our forests properly. Planting of plants is called afforestation. We can plant saplings or barren land or non-agricultural land without converting them into forests. You can improve the deteriorating state of forests by planting 311 social forests, agricultural forests, commercial forests. The government should make every effort to awaken the people regarding the importance of forests and cooperation for their protection. Laws should be strictly enforced by the Forest Department and punish the culprits for cutting down trees, either illegally or without adopting the right method.(A) What are the benefits of forests?(B) How can we save the dwindling forests?

Answer»

(A)

Forests control the climate compact, forests prevent temperature rise in summer and increase the temperature in winter.
The roots of dense forests help in reducing the speed of flowing water, thus reducing the speed of flowing water, thus reducing the outbreak of floods.
The roots of trees strengthen the soil tightness and prevent soil erosion.
The soil gets green manure in the form of bacteria due to dry leaves falling.

(B)

We can imporove the deteriorating state of forests continuously by dividing forests into different parts.
People should be awakened from time to time for the benefits and care of the forests.
Cutting of forests should be declared as a legal offence and strict punishment should be given to those who cut down trees without proper law.
People should be constantly encouraged to plant new trees.

Discussion

32.	Human life is greatly influenced by climate. Man is completely subject to climate at every stage of life, including food, clothes, design of homes, health work, employment etc. In cold climate regions, hot things are used in food such as tea, coffee and hot food. Colder substances like lassi, ice, sorket etc. are used more, in countries with hot climate. Warm clothes such as coats, sweaters (Woollen garments), jackets, blankets etc. are used in cold countries. Homes are open and ventilated in areas with hot climates. The roofs of houses in hilly areas are sloping. Fruits like apples, almonds, cherries in cold winter. Climate and crops of sugarcane, cotton, rice, jute etc. are grown in hot climatic regions.(A) How is human life affected by climate?(B) Which crops grow in cold and hot climate areas?
Answer» (A) Human life is greatly influenced by climate such as: Our food habits are affected by climate such as hot things are eaten in cold areas, Our clothes are also affected by climate such as wearing warm clothes in cold areas, The structure and size of our houses are also affected by climate such that houses in hot areas are open and ventilated, People living in cold and mountainous areas are physically stronger. (B) Fruits like apple, almond, cherry are grown in cold climate areas. Crops like cotton, sugarcane, rice, wheat, jute etc. are grown in areas with hot climate.

32.

Human life is greatly influenced by climate. Man is completely subject to climate at every stage of life, including food, clothes, design of homes, health work, employment etc. In cold climate regions, hot things are used in food such as tea, coffee and hot food. Colder substances like lassi, ice, sorket etc. are used more, in countries with hot climate. Warm clothes such as coats, sweaters (Woollen garments), jackets, blankets etc. are used in cold countries. Homes are open and ventilated in areas with hot climates. The roofs of houses in hilly areas are sloping. Fruits like apples, almonds, cherries in cold winter. Climate and crops of sugarcane, cotton, rice, jute etc. are grown in hot climatic regions.(A) How is human life affected by climate?(B) Which crops grow in cold and hot climate areas?

Answer»

(A) Human life is greatly influenced by climate such as:

Our food habits are affected by climate such as hot things are eaten in cold areas,
Our clothes are also affected by climate such as wearing warm clothes in cold areas,
The structure and size of our houses are also affected by climate such that houses in hot areas are open and ventilated,
People living in cold and mountainous areas are physically stronger.

(B) Fruits like apple, almond, cherry are grown in cold climate areas.

Crops like cotton, sugarcane, rice, wheat, jute etc. are grown in areas with hot climate.

Discussion

33.	A force given by the relation F = 8t, acts on a body of mass 2 kg, initially at rest. Find the work done by this force on the body during first 2 seconds of its motionA. 64 JB. zeroC. `- 64 J`D. none of these
Answer» Correct Answer - A `F = 8t` `m (dv)/(dt) = 8t` `m underset(0)overset(v)int dv = 8 underset(0)overset(t)int t dt` `mv 4t^(2)` or `m = (dx)/(dt) = 4 t^(2)` `rArr dx = (4 t^(2))/(m) dt = 2t^(2) dt` `W = int Fdx = underset(0)overset(v)int (8 t xx 2 t^(2)) dt = 64 J`

Discussion

34.

Explain the rules for counting significant figures with examples?

Answer»

Rules for counting significant figures:

	Rule	Exmaple
(i)	All non- Zero digits are significant	1340 has four significant figures
(ii)	All zeros between two non- zero digits are significant	2008 has four significant figures
(iii)	All zeros to the right of a non- zero digit but to the left of a decimal point are significant.	30700 has three significant figures
(iv)	(a) The number without a decimal point, the terminal or trailing zero (s) are not significant. (b)All zeros are significant if they come from a measurement	(a) 30700 has three significant figures (b) 30700 m has five significant figures
(v)	If the number is less then 1, Zero (s) on the right of the decimal point but to left of the first non- zero digit are snot significant.	0.0035 has three significant figures
(vi)	All zeros to the right of a decimal point and to the right of non-zero digit are significant.	40.00 has four significant figures and 0.030400 has five significant figures
(vii)	The number of significant figures does not depend on the system of units used	1.53 cm, 0.0153 m, 0.0000153 km, all have three significant figures

Discussion

35.	A force given by the relation F = 8t, (F in Newton and t in sec.) acts on a body of mass 2kg. Initially at rest. Choose the correct option.A. Power of force is continuous increasing.B. The work done by this force on the body during first 2 seconds of its motion is 64 J.C. Velocity at 2 seconds after beginning of motion is 8 m/s.D. Power of force is constant.
Answer» Correct Answer - A::B::C Let the both blocks are moving together, then `a = (60)/(40) = 1.5 m//s^(2)` thus, force on lower block has to be F = Ma = `30 xx 1.5 = 45N` Next `f_(L) = mu N = 0.5 xx 10 xx 10 = 50 N` Since `F lt f_(L)` , both the blocks will move together, thus force of friction = 45N

Discussion

36.	A particle is moving along the path given by `y=(C )/(6)t^(6)` (where `C` is a positive constant). The relation between the acceleration (`a`) and the velocity (`v`) of the particle at `t=5sec` isA. 5a = vB. a = 5vC. `a = sqrt(v)`D. a = v
Answer» Correct Answer - A Angular impulse (about com) = `J (L)/(2) = (MV(L)/(2))=(l_(cm))omega=((2ML^(2))/(4))omega`

Discussion

37.	Two block of same mass m undergo headon collision as shown. E is the coefficient of restitution. Choose the correct statement(s) : A. If e = 1, just after collision KE of right block is `(1)/(2)mv^(2)`.B. If e = 0, loss in KE is `(1)/(2)((mv^(2))/(2))`.C. If e = `(1)/(2)`, loss in KE = `(3)/(16)mv^(2)`.D. No impulsive force acts on left block during collision.
Answer» Correct Answer - A::B::D With respect to box, psudo force = mg act upwards. So `F_("net")=0`

Discussion

38.	The number of significant figures in 11.118 × 10-6 is (a) 3 (b) 6 (c) 5 (d) 4
Answer» Correct answer is (C) 5. As per rules number of significant figures in 11.118 × 10^-6 is 5.

38.

The number of significant figures in 11.118 × 10-6 is (a) 3 (b) 6 (c) 5 (d) 4

Answer»

Correct answer is (C) 5.

As per rules number of significant figures in 11.118 × 10^-6 is 5.

Discussion

39.	Fill in the blanks.1. The curved surface area of a solid cylinder of radius 2 cm and height 20 cm is_____m2 (Write answer in 3 significant digits)2. Im = ______ ly
Answer» 1. Curved area = 2πl = 2 × 3.14 (2 × 10²) × 20 × 10² = 2.51 × 10^-6m² 2. l ly= 9.46 × 10¹⁵ m lm = \(\frac{lIy}{9.46×101^5}\) ≈ 10^-6ly.

39.

Fill in the blanks.1. The curved surface area of a solid cylinder of radius 2 cm and height 20 cm is_m2 (Write answer in 3 significant digits)2. Im = __ ly

Answer»

1. Curved area = 2πl
= 2 × 3.14 (2 × 10²) × 20 × 10²
= 2.51 × 10^-6m²

2. l ly= 9.46 × 10¹⁵ m
lm = \(\frac{lIy}{9.46×101^5}\) ≈ 10^-6ly.

Discussion

40.	In a particular experiment Ramu used the relation F = AB + (P + Q) Y to calculate force.1. Which principle is used to check the correctness of the equation (1)2. If the dimensional formula of Y is M0L1T-1, then find the dimensional formula of P
Answer» 1. Principle of homogenity 2. F = AB + (P+Q)Y F = AB + PY + QY MLT^-2 = AB + PY+ QY According to principle of homogeneity MLT^-2 = PY M¹L¹T^-2 = P M⁰L¹T^-1 ie. P = \(\frac{M' L^1 T^{-2}}{M^0L^1Y^{-1}}\)⁼M¹T^-3

40.

In a particular experiment Ramu used the relation F = AB + (P + Q) Y to calculate force.1. Which principle is used to check the correctness of the equation (1)2. If the dimensional formula of Y is M0L1T-1, then find the dimensional formula of P

Answer»

1. Principle of homogenity

2. F = AB + (P+Q)Y

F = AB + PY + QY

MLT^-2 = AB + PY+ QY

According to principle of homogeneity

MLT^-2 = PY

M¹L¹T^-2 = P M⁰L¹T^-1

ie. P = \(\frac{M' L^1 T^{-2}}{M^0L^1Y^{-1}}\)⁼M¹T^-3

Discussion

41.	In an experiment with common balance the mass of a body is found to 2.52g, 2.53g, 2,51g, 2.49g and 2.54g in successive measurements. Calculate1. The mean value of the body2. Mean absolute error3. Percentage error
Answer» 1. Mean value, M_mean = \(\frac{2.52+2.53+2.51+2.49+2.54}{5}\) = 2.5g 2. Absolute error, Absolute error ∆m₁ = \|2.52 – 2.52\| = 0 ∆m₂ = \|2.52 – 2.53\| = 0.01 ∆m₃ = \|2.52 – 2.51\| = 0.01 ∆m₄ = \|2.52 – 2.49\| = 0.03 ∆m₅ = \|2.52 – 2.54\| = 0.02 ∴ Mean absolute error \(\frac{0+0.01+0.01+0.03+0.02}{5}\) ∆m_mean = 0.014g 3. Percentage error = \(\frac{Δm_{mean}}{m_{mean}}\) × 100 = \(\frac{ 0.014}{2.52}\) × 100 = 0.556.

41.

In an experiment with common balance the mass of a body is found to 2.52g, 2.53g, 2,51g, 2.49g and 2.54g in successive measurements. Calculate1. The mean value of the body2. Mean absolute error3. Percentage error

Answer»

1. Mean value, M_mean

= \(\frac{2.52+2.53+2.51+2.49+2.54}{5}\)

= 2.5g

2. Absolute error,

Absolute error ∆m₁ = |2.52 – 2.52| = 0

∆m₂ = |2.52 – 2.53| = 0.01

∆m₃ = |2.52 – 2.51| = 0.01

∆m₄ = |2.52 – 2.49| = 0.03

∆m₅ = |2.52 – 2.54| = 0.02

∴ Mean absolute error

\(\frac{0+0.01+0.01+0.03+0.02}{5}\)

∆m_mean = 0.014g

3. Percentage error = \(\frac{Δm_{mean}}{m_{mean}}\) × 100

= \(\frac{ 0.014}{2.52}\) × 100 = 0.556.

Discussion

42.	Fill in the blanks by suitable conversion of units:1. 1 kgm2s-2 = _____g cm2s-22. 1 m =_____1 y3. 3.0 ms2 =______kmh-24. G = 6.67 × 10-11 Nm2 (kg)-2 =_____(cm)3s-2g-1
Answer» 1. 1 kgm²s^-2 = 10⁷ g cm²s^-2 2. 1 m = 10^-16 1 y 3. 3.0 ms² =3.888 x \(10^4\) kmh^-2 4. G = 6.67 × 10^-11 Nm² (kg)^-2 = 6.67 x 10^-8(cm)³s^-2g^-1

42.

Fill in the blanks by suitable conversion of units:1. 1 kgm2s-2 = ___g cm2s-22. 1 m =_1 y3. 3.0 ms2 =kmh-24. G = 6.67 × 10-11 Nm2 (kg)-2 =_(cm)3s-2g-1

Answer»

1. 1 kgm²s^-2 = 10⁷ g cm²s^-2

2. 1 m = 10^-16 1 y

3. 3.0 ms² =3.888 x \(10^4\) kmh^-2

4. G = 6.67 × 10^-11 Nm² (kg)^-2 = 6.67 x 10^-8(cm)³s^-2g^-1

Discussion

43.	Which of the following is preciseA vernier calliperse with 40 divisions on sliding scaleAn optical instrument that can measure length of the order of wavelength of light.Is it possible to increase the accuracy of screw gauge by increasing the number of divisions on the head scale?
Answer» 1. (i) L.C of vernier caliperse = 140 = 0.025mm = 0.025 × 10^-3m = 2.5 × 10^-5m. (ii) L.C of optical instrument = 6000A° = 6000 × 10^-10m (Taking λ of visible light = 6000°A)= 6 × 10^-7m 2. Yes. Because L.C proportional to number of division on the headscale. So with the increase in number of divisions, the least count will increase. This leads to increase the accuracy of above screw guage.

43.

Which of the following is preciseA vernier calliperse with 40 divisions on sliding scaleAn optical instrument that can measure length of the order of wavelength of light.Is it possible to increase the accuracy of screw gauge by increasing the number of divisions on the head scale?

Answer»

1. (i) L.C of vernier caliperse = 140 = 0.025mm

= 0.025 × 10^-3m

= 2.5 × 10^-5m.

(ii) L.C of optical instrument = 6000A°

= 6000 × 10^-10m

(Taking λ of visible light = 6000°A)= 6 × 10^-7m

2. Yes. Because L.C proportional to number of division on the headscale. So with the increase in number of divisions, the least count will increase. This leads to increase the accuracy of above screw guage.

Discussion

44.	A radioactive element has rate of disintegration `10,000` disintegrations per minute at a particular instant. After four minutes it becomes `2500` disintegrations per minute. The decay constant per minute isA. `0.2 log_(e)^(2)`B. `0.5 log_(e)^(2)`C. `0.6 log_(e)^(2)`D. `0.8 log_(e)^(2)`
Answer» Correct Answer - B `(N)/(N_(0))=e^(-lambdat)` `therefore (2500)/(10000)=e^(-lambda xx 4)` `therefore 1/4=e^(-4)` `therefore e^(4lambda)=4` `therefore 4lambda=log_(e)4` `therefore 4lambda=log_(e)2^(2)` `therefore 4lambda=2log_(e)2` `therefore lambda=2/4 log_(e)2` `therefore lambda=0.5 log_(e)2`

Discussion

45.	In common emitter transistor, the input resistance is `200Omega` and bad resistance is `40kOmega`. If current gain is `80` then voltage gain isA. `16`B. `160`C. `1600`D. `16000`
Answer» Correct Answer - D `"voltage gain"=("current gain")("resistance gain")`

Discussion

46.	A sound source emits two sinusoidal sound waves, both of wavelength `lambda`, along paths A and B as shown in figure.The sound travelling along path B is reflected from five surfaces as shown and then merges at point Q, producing minimum intensity at that point.The minimum value of d in terms of `lambda` is : A. `lambda/8`B. `lambda/4`C. `(3lambda)/8`D. `lambda/2`
Answer» Path difference = Path B-Path A =7d-3d=4d [Note that there is no phase change in reflections from mirror in case of sound] For being out of phase : `Deltax=4d=lambda/2 , (3lambda)/2`,.... For minimum d, `4d=lambda/2` `rArr d=lambda/8`

Discussion

47.	`S_1 and S_2` are two coherent sources of radiations separated by distance `100.25 lambda`, where `lambda` is the wave length of radiation. `S_1` leads `S_2` in phase by `pi//2`.A and B are two points on the line joining `S_1 and S_2` as shown in figure.The ratio of amplitudes of component waves from source `S_1 and S_2` at A and B are in ratio 1:2. The ratio of intensity at A to that of B `(I_A/I_B)` is A. `oo`B. `1/9`C. 0D. 9
Answer» For interference at A:`S_2` is behind of `S_1` by a distance of `100lambda+lambda/4`.(equal to phase difference `pi/2`).Further `S_2` lags `S_1` by `pi/2` Hence the waves from `S_1 and S_2`interfere at A with a phase difference `200.5pi+0.5pi=201pi=pi` Hence the net amplitude at A is 2a-a=a For interference at B :`S_2` is ahead of `S_1` by a distance of `100lambda+lambda/4`.(equal to phase difference `pi/2`).Further `S_2` lags `S_1` by `pi/2` Hence waves from `S_1 and S_2` interfere at B with a phase difference of `200.5 pi-0.5pi=200pi=0pi`. Hence the net amplitude at A is 2a+a=3a Hence `(I_A/I_B)=(a/(3a))^2=1/9`

Discussion

48.	A transistor is used in common-emitter configuration. Given its `alpha=0.9`. The change in collector current when the base current changes by `2mA` isA. `0.9mA`B. `18mA`C. `20mA`D. `0.1mA`
Answer» Correct Answer - B `alpha=(Deltai_(c))/(Deltai_(e))=0.9=(Deltai_(c))/(Deltai_(c)+Deltai_(b))`

Discussion

49.	An unknown quantity `'alpha'` is expressed as `alpha = (2ma)/(beta) log(1+(2betal)/(ma))` where m = mass, a = acceleration `l` = length, The unit of `alpha` should beA. meterB. m/sC. `m//s^(2)`D. `s^(-1)`
Answer» Correct Answer - A `[(2 beta l)/(ma)] = M^(0)L^(0)T^(0)` `[(beta)/(ma)] = [(1)/(l)]` `[alpha] = [(ma)/(beta)] = [l]`

Discussion

50.	Write the nuclear reactions for the following radioactive decay: (a) `._(92)U^(238)` undergoes `alpha-` decay. (b) `._(91)Pa^(234)` undergoes `Beta-` decay. (c) `._(11)Na^(22)` undergoes `Beta^(+)` decay.
Answer» `a. ` `._(92)U^(238)rarr ` `._(90)Th^(234)+` `._(2)He^(4)` `b.` `._(91)Pa^(234)rarr ` `._(2)U^(234)+` `._(-1)e^(0)(beta^(c-))` `c.` `._(11)Na^(22)rarr ` `._(10)Ne^(22)+` `._(+1)e^(0)(beta^(o+))`

Discussion

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