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The number of ways in which a mixed doubles game in tennis can be arranged from 5 married couples, if no husband and wife play in the same game, is

(a) 46 (b) 54 (c) 60 (d) None of these Solution: Given that there are 5 married couples. Let the sides of the game... View Article

Let 1 ≤ m < n ≤ p. The number of subsets of the set A = {1, 2, 3, ……..p} having m, n as the least and the greatest elements respectively, is

(a) 2n-m-1-1 (b) 2n-m-1 (c) 2n-m (d) 2p-n+m-1 Solution: Given 1 ≤ m < n ≤ p A = {1, 2, 3, ........p} n is the... View Article

n’ is selected from the set {1, 2, 3, …, 100} and the number 2ⁿ + 3ⁿ + 5ⁿ is formed. Total number of ways of selecting ‘n’ so that the formed number is divisible by 4, is equal to

(a) 50 (b) 49 (c) 48 (d) None of these Solution: If n is odd, 3n = 4λ1 - 1 5n = 4λ2 + 1 => 2n... View Article

If 12 persons are seated in a row, the number of ways of selecting 3 persons from them, so that no two of them are seated next to each other is

(a) 85 (b) 100 (c) 120 (d) 240 Solution: The number of ways of selecting 3 persons from 12 people under the given... View Article

All the words that can be formed using alphabets A, H, L, U, and R are written as in a dictionary (no alphabet is repeated). The rank of the word RAHUL is

(a) 71 (b) 72 (c) 73 (d) 74 Solution: No. of words starting with A are 4 ! = 24 No. of words starting with H are 4... View Article

A boat is to be manned by eight men of whom 2 can only row on the bow side and 1 can only row on stroke side, the number of ways in which the crew can be arranged is

(a) 4360 (b) 5760 (c) 5930 (d) None of these Solution: 2 men on the bow side can be arranged in 4P2 ways = 12 1 man on... View Article

A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is

(a) 10!/6! (b) 10!/24 (c) 9!/24 (d) none of these Solution: 6 friends are at one round table and the remaining 4 are at the other round... View Article

A student performed the experiment to measure the speed of sound in air using the resonance air-column method. Two resonances in the air column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air-column is the second resonance. Then,

a) the intensity of the sound heard at the first resonance was more than that at the resonance b) the prongs of the tuning fork were kept in a... View Article

The frequency of the fundamental note in a wire stretched under tension T is v. If the tension is increased to 25 T, then the frequency of the fundamental note will be

a) 25f b) 5f c) 10f d) f Answer: b) 5f Solution: The fundamental frequency f = (1/2L) √(T/m) (f'/f)... View Article

A transverse wave propagating on a stretched string of linear density 3 × 10^-4 kg-m^-1 is represented by the equation, y = 0.2 sin (1.5x + 60t ), where, x is in metre and t is in second. The tension in the string (in newton) is

a) 0.24 b) 0.48 c) 1.20 d) 1.80 Answer: b) 0.48 Solution: y = 0.2 sin (1.5x + 60t ) Comparing with standard... View Article

Two closed organ pipe A and B have the same length. A is wider than B. They resonate in the fundamental mode at frequencies v_A and v_B respectively, then

a) vA = vB b) vA > vB c) vA < vB d) either (2) or (3) depending on the ratio of their diameters Answer: c) vA < vB Solution: In... View Article

In an experiment with a sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. The tension of the string remains constant. The frequency of the second tuning fork is

a) 163.84 Hz b) 400 Hz c) 320 Hz d) 204.8 Hz Answer: b) 400 Hz Solution: F = (1/2L) √(T/m) F ∝ (1/L) F... View Article

In Melde’s experiment in the transverse mode, the frequency of the tuning fork and the frequency of the waves in the string are in the ratio

a) 2 : 1 b) 4 : 1 c) 1 : 1 d) 1 : 2 Answer: 1:1 Solution: The tuning fork and the string are in resonance So, their frequencies are equal... View Article

A segment of wire vibrates with a fundamental frequency of 450 Hz under a tension of 9 kg–wt. Then, tension at which the fundamental frequency of the same wire becomes 900 Hz is

a) 36 kg–wt b) 27 kg–wt c) 18 kg–wt d) 72 kg–wt Answer: a) 36 kg–wt Solution: Frequency,... View Article

A wave has velocity v in medium P and velocity 2v in medium Q. If the wave is incident in medium P at an angle of 30⁰, then the angle of refraction will be

a) 300 b) 450 c) 600 d) 900 Answer: d) 900 Solution: Given, A wave has velocity v in medium P and velocity 2v in medium Q... View Article

Two periodic waves of intensities I₁ and I₂ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

a) I1 + I2 b)(√I1 +√I2)2 c) (√I1 - √I2)2 d) 2(I1 + I2) Answer: d) 2(I1 + I2) If two... View Article

When both the listener and source are moving towards each other, then which of the following is true regarding frequency and wavelength of the wave observed by the observer?

a) More frequency, less wavelength b) More frequency, more wavelength c) Less frequency, less wavelength d) More frequency, constant... View Article

Radar waves are sent towards a moving aeroplane and the reflected waves are received. When the aeroplane is moving towards the radar, the wavelength of the wave

a) decreases b) increases c) remains the same d) sometimes increases or decreases Answer: a) decreases Solution: When an... View Article

The extension in a string obeying Hooke’s law is x. The speed of sound in the stretched string is v. If the extension in the string is increased to 1.5 x, the speed of sound will be

a) 1.22 v b) 0.61 v c) 1.50 v d) 0.75 v Answer: a) 1.22 v Solution: Velocity of sound by a stretched string, v =... View Article

A glass tube of length 1.0 m is completely filled with water. A vibrating tuning fork of frequency 500 Hz is kept over the mouth of the tube and the water is drained out slowly at the bottom of the tube. If the velocity of sound in air is 330 ms ^-1, then the total number of resonances, that occur will be

a) 2 b) 3 c) 1 d) 5 Answer: b) 3 Solution: f=v/4l ⇒ l=v/4f Therefore, l1=v/4f1 =330/(4×500) =0.165m... View Article

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