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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Consider an arithmetic sequence whose 7th term is 34 and 15th term is 66.a. Find the common difference,b. Find the 20th term. |
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Answer» a. 15th term can be obtained by adding 7th term ‘ and 8 times the common difference. x15 = x7 + 8d 66 = 34 + 8d 8d = 66 – 34 = 32 d = \(\frac{32}{8}\) = 4 d. 20th term can be obtained by adding 15th term and 5 times the common difference. x20 = x15 + 8d = 66 + 5 × 4 = 86 |
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| 2. |
If the angles of a right triangle are in an arithmetic sequence, find them by making suitable equations. |
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Answer» Let angles be f – d, f, f + d f – d + f + f + d = 180 3f = 180, f = 60, f + d = 90, d = 30 Angles are 30, 60 and 90 |
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| 3. |
Write eighth terms of an arithmetic sequence using the numbers given below. (22, 15, 18, 4, 10, 14, 6, 12) |
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Answer» Answer is 6, 10, 14, 18, 22, …………… |
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| 4. |
Write the algebraic form of 1,4,7,10,… is 100 a term of this sequence. Why? Prove that the square of any term of this sequence belongs to that sequence. |
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Answer» 1,4, 7, 10,… f = 1, d = 3 xn = dn + (f – d) = 3n + (1 – 3) = 3n – 2 When 4 is divided by 3 we get remainder as 1 When 100 is divided by 3 we get remainder as 1 So, 100 is a term of the arithmetic sequence. Square = (3n – 2)2 = 9n2 – 12n + 4 (9n2 – 12n + 4) + 3 When 9n2 – 12n + 4 is divided by 3the remainder is 1. So square of the term also in the sequence. ∴ The square of any term of this sequence belongs to that sequence. |
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| 5. |
Write the sequence of natural numbers |
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Answer» Answer is 1,2,3,4, |
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| 6. |
Write the sequence obtained by adding two adjacent consecutive terms in counting numbers starting from 1. Write the algebraic expression of this sequence. |
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Answer» Counting numbers: 1, 2, 3, 4, 5, … Sequence obtained by adding: : 1 + 2, 2 + 3, 3 + 4, 4 + 5, … Two adjacent consecutive terms 3, 5, 7, 9, … Algebraic expression of above sequence : n + (n + l) = 2n + 1 |
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| 7. |
The algebraic form of an arithmetic sequence is 3 + 2n. a What is the first form of the sequence? b. What will be the remainder if the terms of the sequences are divided by 2? |
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Answer» a. Firstterm = 3 + 2 x 1 = 5 b. d = 2 (coefficient of n be the common difference) The remainder divided by 2 = 1 |
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| 8. |
the sum of in coming number is n (n + 1)by 2a) find the sum of first 25 counting numbersb) find the sum of first 25 even numbersc) find the sum of first 25 odd numbers |
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Answer» a) Given an array and an integer K, find the maximum for each and every contiguous subarray of size k. Examples : Input: arr[] = {1, 2, 3, 1, 4, 5, 2, 3, 6}, K = 3 Output: 3 3 4 5 5 5 6 Explanation: Maximum of 1, 2, 3 is 3 Maximum of 2, 3, 1 is 3 Maximum of 3, 1, 4 is 4 Maximum of 1, 4, 5 is 5 Maximum of 4, 5, 2 is 5 Maximum of 5, 2, 3 is 5 Maximum of 2, 3, 6 is 6 Input: arr[] = {8, 5, 10, 7, 9, 4, 15, 12, 90, 13}, K = 4 Output: 10 10 10 15 15 90 90 Explanation: Maximum of first 4 elements is 10, similarly for next 4 elements (i.e from index 1 to 4) is 10, So the sequence generated is 10 10 10 15 15 90 90 Recommended: Please solve it on “PRACTICE” first, before moving on to the solution. Method 1: This is a simple method to solve the above problem. Approach: The idea is very basic run a nested loop, the outer loop which will mark the starting point of the subarray of length k, the inner loop will run from the starting index to index+k, k elements from starting index and print the maximum element among these k elements. Algorithm: Create a nested loop, the outer loop from starting index to n – k th elements. The inner loop will run for k iterations. Create a variable to store the maximum of k elements traversed by the inner loop. Find the maximum of k elements traversed by the inner loop. Print the maximum element in every iteration of outer loop b) How to Find the Sum of First 25 Even Numbers? The below workout with step by step calculation shows how to find what is the sum of first 25 even numbers by applying arithmetic progression. It's one of the easiest methods to quickly find the sum of given number series. step 1 Address the formula, input parameters & values. Input parameters & values: The number series 2, 4, 6, 8, 10, 12, . . . . , 50. The first term a = 2 The common difference d = 2 Total number of terms n = 25 step 2 apply the input parameter values in the AP formula Sum = n/2 x (a + Tn) = 25/2 x (2 + 50) = (25 x 52)/ 2 = 1300/2 2 + 4 + 6 + 8 + 10 + 12 + . . . . + 50 = 650 Therefore, 650 is the sum of first 25 even numbers. c) The odd numbers are: 1, 3, 5, 7, 9,….49 The first term is = 1 The common difference is, d = 2 The total no.of terms is, n = 25 Sum = n/2 x (a+Tn) =25/2 x (1+49) =1250/2 = 625 Therefore, the sum of first 25 odd numbers is 625. |
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| 9. |
Can the difference between any two terms of an arithemetic sequence having common difference 6 be 2016? Justify your answer. |
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Answer» \(\frac{Differences\,of\,terms}{Common\,difference}=\frac{2016}{6}=336\) It is a natural number, so the difference between any two terms of an arithemetic sequence having common difference 6 be 2016. |
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| 10. |
98 is a term of the arithmetic sequence having common difference 7. is 2016 a term of this sequence. Why? |
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Answer» If (2016 – 98) is a multiple of common difference 7, then 2016 is a term of the arithmetic sequence. 2016 – 98 = 1918 1918 is a multiple of 7. (Quotient = 274, Remainder = 0) ∴ 2016 is a term of this sequence. |
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| 11. |
When 60 added to the first term of an A.P, we get its 11th term. Which number should be added to its first term to get the 19th term? Can 75 be the difference between any of the two terms of this sequence? |
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Answer» Differences between first term and 11th term is = 60 (11 – 1 = 10) 10 times of common difference = 60 Common difference = 60/10 = 6 Differences between first term and 19th term is = 18 x common difference = 18 x 6 = 108 When 108 is added to the first term, we get 19th term. The difference between two terms in an A.P is the multiple of common difference. 75 is not the multiple of common difference 6. So 75 cannot be the difference between two terms in the series. |
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| 12. |
—, 18, —, 28 are four consecutive terms of an arithmetic sequence. Fill in the blanks. |
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Answer» 18 + 2d = 28 2d = 28 – 18 = 10 d = 5 Sequence 13, 18, 23, 28. |
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| 13. |
In an arithmetic sequence having terms natural numbers, prove that if one of the terms is a perfect square, it will have more that this as the perfect square term. |
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Answer» As we know when a definite number of common difference of an arithmetic sequence is added to a term we get another term of the same sequence. If n2 is a perfect square term, add (2n + d) times d to n2. n2 + (2n + d) x d = (n + d)2. This is nothing but a perfect square term. |
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| 14. |
What number should be added with 5474 so that the sum will be completely divisible by 3, 4, 6 and 8?1. 22. 223. 184. 12 |
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Answer» Correct Answer - Option 2 : 22 Given: The number will be completely divisible by 3, 4, 6 and 8 Calculation: LCM of 3, 4, 6 and 8 is 24 Now on dividing 5474 by 24 we get the remainder = 2 To get the desired number we have to add (24 - 2) = 22 ∴ 22 should be added with 5474 so that the sum will be completely divisible by 3, 4, 6 and 8. |
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| 15. |
l ϕf is the phase difference between ac current and e.m.f., then the value of power factor is(a) tan ϕ(b) cos2 ϕ(c) sin ϕ(d) cos ϕ |
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Answer» Correct answer is (d) cos ϕ |
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| 16. |
The colour of a thin film is due to(a) scattering(b) interference(c) dispersion(d) diffraction |
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Answer» (b) interference |
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| 17. |
The colour of a thin film is due to(a) scattering(b) interference(c) dispersion(d) diffraction |
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Answer» (b) interference |
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| 18. |
The transerse nature of light wave supports(a) interference(b) reflection (c) polarisation(d) dispersion |
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Answer» (a) interference |
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| 19. |
When two converging lenses of same focal length f are placed in contact, then the focal length of the combination is(a) f(b) 2f(c) f/2 (d) 3f. |
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Answer» Correct answer is (b) 2f |
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| 20. |
The final image in astronomical telescope is(a) real and erect(b) real and inverted(c) virtual and inverted(d) virtual and erect. |
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Answer» (c) virtual and inverted |
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| 21. |
In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fifth digit is the sum of first two digits, the third digit is equal to the first digit, the second digit is twice the first digit and the fourth digit is the sum of fifth and sixth digits. Then, the largest possible value of the fourth digit is (type in box) |
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Answer» Calculation: Let the first digit be 'a' Now according to the question, Third digit = The first digit ⇒ a The second digit = 2 × The first digit ⇒ 2 × a ⇒ 2a The sixth digit = The sum of (first digit + second digit + third digit) ⇒ a + 2a + a ⇒ 4a The fifth digit = The sum of first two digit ⇒ a + 2a ⇒ 3a The fourth digit = The sum of (fifth digit + sixth digit) ⇒ 3a + 4a ⇒ 7a So we can see our fourth digit should be a multiple of 7 And we can say that the largest possible value of the fourth digit will be 7 when the value of 'a' will be 1 Because if we put 'a' equal to 2 it will give the two-digit number equal to (7 × 2 = 14) for the fourth digit which is not possible. ∴ The largest possible value of the fourth digit is only 7. |
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| 22. |
Power of electric circuit is(a) V .R(b) V2 .R(c) V2/R(d) V2.R.l |
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Answer» Correct answer is (c) V2/R |
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| 23. |
Lenz's law is associated with(a) charge(b) mass(c) energy(d) principle of conservation of momentum |
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Answer» (d) principle of conservation of momentum. |
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| 24. |
A six-digit number 4ABB8A is a multiple of 33 for non-zero digits A and B. Which of the following could be the possible values of A and B?1. A = 3, B = 22. A = 6, B = 33. A = 6, B = 24. A = 3, B = 6 |
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Answer» Correct Answer - Option 2 : A = 6, B = 3 Given: Number = 4ABB8A Concept used: 33 = 3 × 11 So, the divisibility rule of 3 and 11 must be satisfied for the number to be multiple of 3 and 11. Divisibility rule of 3: The sum of the digits must be a multiple of 3. Divisibility rule of 11: Difference between the sum of alternate digits of the number = 0 or multiple of 11 Calculation: Applying divisibility rule of 3: 4 + A + B + B + 8 + A = 12 + 2B + 2A = 12 + 2(A + B) For the number to be multiple of 3, A + B must be multiple of 3 according to the above sum. This is satisfied by options 2 and 4 only. Applying divisibility rule of 11: Sum of alternate digits = (4 + B + 8) and (A + B + A) Difference between the sum = 12 + B - (2A + B) = 11N (N = 0, 1, 2, ....) 12 - 2A = 0 (least multiple) Then, A = 6 Then, from the options 2 and 4, only option 2 is possible. ∴ The possible option is 2 only. |
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| 25. |
Electron-volt (eV) is the measure of(a) charge(b) potential difference(c) current(d) energy. |
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Answer» Correct answer is (d) energy |
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| 26. |
Electron-volt (eV) is the measure of:(a) charge (b) potential difference(c) current(d) energy |
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Answer» Correct answer is (d) energy |
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| 27. |
The working of dynamo is based on the principle of :(a) heating effect of current(b) electromagnetic induction(c) magnetic induction(d) electric induction |
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Answer» (b) electromagnetic induction |
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| 28. |
Two coherent waves represented by `y_(1) = A sin ((2 pi)/(lambda) x_(1) - omega t + (pi)/(6))` and `y_(2) = A sin (( 2pi)/(lambda) x_(2) - omega t + (pi)/(6))` are superposed. The two waves will produceA. constructive interference at `(x_(1)-x_(2))=2lambda`B. constructive interference at `(x_(1)-x_(2))=23/24 lambda`C. destructive interference at `(x_(1)-x_(2))=1.5lambda`D. destructive interference at `(x_(1)-x_(2))=11/24 lambda` |
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Answer» Correct Answer - B::D `Deltaphi=(2pi)/(lambda)(x_(1)-x_(2))+(pi/4-pi/6)` when `x_(1)-x_(2)=23lambda//24` then`Deltaphi=2pi`(constructive interference) when`x_(1)-x_(2)=11lambda//24` then`Deltaphi=pi`(destructive interference) |
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| 29. |
Consider an FM wave f(t) = cos[2πfct + \(\beta\)1 sin 2πf1t + \(\beta\)22πf2t] . The maximum deviation of the instantaneous frequency from the carrier frequency fc is (a) \(\beta\)1f1 + \(\beta\)2f2 (b) \(\beta\)1f2+ \(\beta\)2f1 (c) \(\beta\)1 + \(\beta\)2 (d) f1 + f2 |
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Answer» (a) \(\beta\)1f1 + \(\beta\)2f2 The instantaneous value of the angular frequency ωi = ωc + d/dt(\(\beta\)1 sin 2πf1t + \(\beta\)22πf2t) ωc +\(\beta\)12πf1 cos 2πf1t + \(\beta\)22πf2 cos 2πf2t fi = fc + \(\beta\)1f1 cos 2πf1t + \(\beta\)2f2 cos 2πf2t |
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| 30. |
A modulation signal is y(t) = m(t) cos(40000t), where the baseband signal m(t) has frequency components less than 5 kHz only. The minimum required rate (in kHz) at which y(t) should be sampled to recover m(t)is |
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Answer» The minimum sampling rate is twice the maximum frequency called Nyquist rate The minimum sampling rate (Nyquist rate) = 10K samples/sec. |
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| 31. |
Consider an angle modulation signal x(t) = 6cos[2 × 103 + 2 sin(8000t) + 4 cos(8000t)]. The average power of x(t) is (a) 10 W (b) 18 W (c) 20 W (d) 28 W |
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Answer» (b) 18 W The average power of an angle modulated signal is Ac 2/2 = 62/2 = 18 W |
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| 32. |
A person takes `1//2` kg of cheese sandwitches of energy equivalent to `813 kJ`. Suppose that all the energy is lost only through perspiration, what mass of water would he need to perpire in order to maintain his original temperature. Given : Enthalpy of vapurisation of water is `40.65 kJ mol^(-1)`.A. 360gB. 3.6 gC. 180gD. 190g |
| Answer» Correct Answer - A | |
| 33. |
Select the option having all extensive terms.A. pH, electrode potential, molesB. Specific heat capacity, resistance, molar enthalpyC. Heat capacity, resistance, enthalpyD. Mass, molarity, resistivity |
| Answer» Correct Answer - C | |
| 34. |
The tail current equals the a. Difference between two emitter currents b. Sum of two emitter currents c. Collector current divided by current gain d. Collector voltage divided by collector resistance |
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Answer» (b) Sum of two emitter currents |
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| 35. |
A 10 MHz carrier is frequency modulated by a sinusoidal signal of 500 Hz, the maximum frequency deviation being 50 KHz. The bandwidth required. as given by the Carson’s rule is........... |
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Answer» By carson’s rule BW = 2(Δf + fm) = 2(50+ 0.5) = 101 KHz |
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| 36. |
A dc signal has a frequency of a. 0 b. 60 Hz c. 0 to over 1 MHz d. 1 MHz |
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Answer» The correct answer is: (a) 0 |
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| 37. |
The voltage gain of a diff amp with a differential output is equal to RC divided by a. re' b. \(\frac{re'}{2}\) c. 2re' d. RE |
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Answer» The correct answer is: (a) re' |
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| 38. |
The common-mode voltage gain of a diff amp is equal to RC divided by a. re' b. \(\frac{re'}{2}\) c. 2re' d. 2RE |
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Answer» The correct answer is: (d) 2RE |
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| 39. |
The stationary wave`y = 2a sin kx cos omega t` in a closed organ pipe is the result of the superposition of `y = a sin (omega t - kx)`A. `y = a cos(omegat- kx)`B. `y = -a sin(omegat+ kx)`C. `y = a sin(omegat- kx)`D. `y = a cos(omegat+ kx)` |
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Answer» Correct Answer - D Potential across the capacitor is zero, hence energy stored is zero. |
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| 40. |
One end of thin aluminium rod (cross -sectional Area `=10^(-6)m^(2)`) is bult welded to the end of copper rod with same diameter. Both rods are very long and under uniform tension of `900N` Given that `Y_((Cu))=1.6xx10^(11)Nm^(-2), Y_((Al))=0.9xx10^(11)Nm^(-2)` `rho_((Cu))=8.1xx10^(3)kgm^(-3), rho_((Al))=2.5xx10^(3)Kgm^(-3)`A. The value of relection factor `(A_(r))/(A_(i))` for longitudinal pulse approacing the junction along copper rod is 0.15B. The value of reflection factor `(A_(r))/(A_(i))` for longitudinal pulse approachig the junction along copper rod is approximately 0.3C. The value of reflection factor `(A_(r))/(A_(i))` or transverse pulse approaching the junction along copper rod is `0.15`D. The value of reflection factor `(A_(r))/(A_(i))` for transverse pulse approaching the junction along copper rod is approximately 0.3 |
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Answer» Correct Answer - B::C For transverse wave, `v_(T)=sqrt(F/(mu))=sqrt(F/(rhoxxS))` `V_(Al)=sqrt(900/(2.5xx10^(3)xx10^(-6)))=30/(50xx10^(-3))=600m//s` `V_(Cu)=sqrt(900/(8.1xx10^(3)xx10^(-6)))=30/(90xx10^(3))=1000/3m//s` `(a_(r))/(a_(i))=(V_(Al)-V_(Cu))/(V_(Al)+(V_(Cu)))=(800/3)/(2800/3)=2/7~~0.3` For longitudinal wave `V_(l)=sqrt(Y/(rho))` `V_(Al)=sqrt((9xx10^(10))/(2.5xx10^(3)))=(3xx10^(5))/50=6000m//s` `V_(Cu)=sqrt((16xx10^(10))/(8.1xx10^(3)))=(4xx10^(5))/(9xx10)=40000/9m//s` `(a_(r))/(a_(i))=(V_(Al)-V_(Cu))/(V_(Al)+V_(Cu))=(1400/9)/(94000/9)=7/47~~0.15` |
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| 41. |
Consider a wave sent down a string in the positive direction whose equation is given is `y=y_(0)sin[omega(t-x/v)]` The wave is propagated along because each string segment pulls upward and downward on the segment adjacent to it a slightly larger value of `x` and, as a result does work upon the string segment to which wave is travelling. For example, the portion of string at point `A` is going upward, and will pull the portion at point `B` upward as well. In fact, at any point along the string, each segment of the string is pulling on the segment just adjacent and to its right, causing the wagve to propagate. It is by this process that the energy is sent along the string. Now we try to calculate how much energy is propagated down the string per second `T_(y)=tsintheta~~TtanthetaimpliesT_(y)=-T(dely)/(delx)` (The negative sign appears because as shown in the figure II, the slope is negative) the force will act through a distance `dy=v_(y)dt=(dely)/(delt) dt` Therefore work done by force in time `dt` is `dW=T_(y)dy=-T((dely)/(delx))((dely)/(delt))dtimpliesdW=(omega^(2)y_(0)^(2)T)/vcos^(2)[omega(t-x/V)]dt`.....(A) How much average power is transmitted dawn a string having density `mu=5xx10^(-4)kg//m` and `T=5N` by a `200Hz` vibration of amplitude of `0.20cm`A. `0.08` WattB. `0.16` WattC. `0.32` WattD. `0.64` Watt |
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Answer» Correct Answer - B Since the work done will be same at any place along the string in one cycle so choose `x=0`, for convenience. Then work done in a cycle `=(omega^(2)y_(0)^(2)T)/v int_(0)^(2pi//omega) cos^(2) omegat dt=(piomegay_(0)^(2)T)/v` Average power `=` work done per unit time `=("work done in a cycle")/("Time period")=((piomega_(0)^(2)T).v)/((2pi)/omega)=(omega^(2)y_(0)^(2)T)/(2V)` Average power `=2pi^(2)f^(2)y_(0)^(2)T sqrt((mu)/T)=2pi^(2)f^(2)y_(0)^(2)sqrt(muT)` `=2xx10xx200xx200xx0.2xx0.2xx10^(-4)sqrt(5xx10^(-4)xx5)=160xx10^(-3)` Watt `=0.16` Watt. |
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| 42. |
Consider a wave sent down a string in the positive direction whose equation is given is `y=y_(0)sin[omega(t-x/v)]` The wave is propagated along because each string segment pulls upward and downward on the segment adjacent to it a slightly larger value of `x` and, as a result does work upon the string segment to which wave is travelling. For example, the portion of string at point `A` is going upward, and will pull the portion at point `B` upward as well. In fact, at any point along the string, each segment of the string is pulling on the segment just adjacent and to its right, causing the wagve to propagate. It is by this process that the energy is sent along the string. Now we try to calculate how much energy is propagated down the stirng per second `T_(y)=tsintheta~~TtanthetaimpliesT_(y)=-T(dely)/(delx)` (The negative sign appears because as shown in the figure II, the slope is negative) the force will act through a distance `dy=v_(y)dt=(dely)/(delt) dt` Therefore work done by force in time `dt` is `dW=T_(y)dy=-T((dely)/(delx))((dely)/(delt))dtimpliesdW=(omega^(2)y_(0)^(2)T)/vcos^(2)[omega(t-x/V)]dt`.....(A) The average power transmitted down the string (i.e. the average energy transfer or sent down it per second) isA. `(2omega^(2)y_(0)^(2)T)/v`B. `(omega^(2)y_(0)^(2)T)/v`C. `(omega^(2)y_(0)^(2)T)/(2v)`D. `(omega^(2)y_(0)^(2)T)/(4v)` |
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Answer» Correct Answer - C During one time period `(T_(0))` `W=(Tomega^(2)y_(0)^(2))/(2v)[int_(0)^(T_(0))(1+cos{2omega(t-x/v)})dt]` Average Power `=W/(T_(0))=(omega^(2)y_(0)^(2)T)/(2V)` |
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| 43. |
Don't forget to explain |
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Answer» Water absorption through roots can be increased by keeping the plant under the fan. Explanation: The rate of transpiration increases in windy atmosphere. When we place a plant under the fan the speed of air flow is very high. Transpiration will take place in presence of high air flow through the stomata. We have studied that increase in the rate of transpiration increases the water absorption also. As transpiration occurs, water emits out of the leaf creating a vacant space which pulls (transpiration pull) the water from the xylem vessels. This results in the absorption of water by the root from the soil. |
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| 44. |
Why mountaineers carry oxygen with them? |
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Answer» Mountaineers carry oxygen with them because there is deficiency of oxygen on mountains at high altitude. |
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| 45. |
Are capillaries joined to veins? If yes then why? |
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Answer» Capillaries are connected to both veins and arteries thus playing the dual role in exchange oxygen as well as carbon dioxide. Capillaries are very delicate, sometimes as thin as one epithelial cell so that blood can easily pass through them. |
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| 46. |
A stone weight 10N falls from the top of a 250 m high cliff. How much energy is transffered to the stone? |
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Answer» Potential energy = Mgh Mg (weight) is given as 10N |
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| 47. |
A ballon has 5.0 g mole of helium at 7°C. Calculate(a) the number of atoms of helium in the balloon(b) the total internal energy of the system. |
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Answer» μ = 5.0 T = 280K No of atoms = μNA = 5.0 x 6.02 x 1023 = 30 × 1023 Average kinetic energy per molecule = (3/2)kT Total internal energy = (3/2)kT x N = (3/2) x 30 x 1023 x 1.38 x 10-23 x 280 = 1.74 x 104 J |
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| 48. |
A silver ball painted black is kept inside a box which is maintained at a temprature of `27^(@)C` the ball is maintained initially at a constant temperature of `127^(@)C` by making the radiation to fall on it through a small hole in the box latter on due to some chemical reaction between silver and paint the paint uniformly evaporates from the surface of ball exposing the silver if same amount of radiation continues to fall on ball then temperature of ball as a function of time is hown as :(Assume emissivity of silver is zero and paint to be black body also assume radiation to be the only mode of heat transfer) A. B. C. D. |
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Answer» Correct Answer - C Rate of absorption = rate of emission `P_(ab)+P_(ab)=P_(cm)` `eAsigma T_(0)^(4)=eAsigmaT_(1)^(4)=eAsigmaT_(B)^(4)` `T_(B)`=remains constant as `(T_(0)` and `T_(1))` are constant |
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| 49. |
Identify the graph which correctly represents the spectral intensity versus wavelength graph at two temperatures T and T `(TleT)`A. B. C. D. none of these |
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Answer» Correct Answer - B As timeperature increases spectral intensity corresponding to all wavelengths increases but the wavelength with maximum spectral intensity decreases according to wein displacement law |
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| 50. |
The crowd expected him to arrive at ten o’clock. |
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Answer» He was expected to come in time by the crowd. |
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