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Physics NEET Questions — Free MCQ Practice

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A string is producing transverse vibration whose equation is $y = 0 .021 \sin (x + 30 t)$ , Where x and y are in meters and t is in seconds. If the linear density of the string is 1.3×10^–4 kg/m, then the tension in the string in N will be :

A.

10

B.

0.5

C.

1

D.

0.117

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Explanation

$y = 0 .021 \sin (x + 30 t)$ ⇒ $v = \frac{ω}{k} = \frac{30}{1} = 30 m / s$ .

Using, $v = \sqrt{\frac{T}{m}} \Rightarrow 30 = \sqrt{\frac{T}{1 .3 \times 10^{- 4}}} \Rightarrow T = 0 .117$ N

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A stretched string of length l, fixed at both ends can sustain stationary waves of wavelength λ, given by

A.

$λ = \frac{n^{2}}{2 l}$

B.

$λ = \frac{l^{2}}{2 n}$

C.

$λ = \frac{2 l}{n}$

D.

$λ = 2 l n$

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Explanation

$λ = \frac{2 l}{p}$ (p = Number of loops)

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A string on a musical instrument is 50 cm long and its fundamental frequency is 270 Hz. If the desired frequency of 1000 Hz is to be produced, the required length of the string is :

A.

13.5 cm

B.

2.7 cm

C.

5.4 cm

D.

10.3 cm

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Explanation

$n \propto \frac{1}{l} \Rightarrow \frac{l_{2}}{l_{1}} = \frac{n_{1}}{n_{2}} \Rightarrow l_{2} = l_{1} (\frac{n_{1}}{n_{2}}) = 50 \times \frac{270}{1000} = 13 .5 c m$

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The tension in a piano wire is 10N. What should be the tension in the wire to produce a note of double the frequency :

A.

5 N

B.

20 N

C.

40 N

D.

80 N

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Explanation

$n \propto \sqrt{T}$ ⇒ $\frac{n_{1}}{n_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$

⇒ $\frac{n}{2 n} = \sqrt{\frac{10}{T_{2}}}$ ⇒ $T_{2} = 40 N$

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A string of 7 m length has a mass of 0.035 kg. If the tension in the string is 60.5 N, then the speed of a wave on the string is :

A.

77 m/s

B.

102 m/s

C.

110 m/s

D.

165 m/s

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Explanation

$v = \sqrt{\frac{T}{m}}$ ⇒ $v = \sqrt{\frac{60 .5}{(0 .035 / 7)}} = 110 m / s$

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A second harmonic has to be generated in a string of length l stretched between two rigid supports. The point where the string has to be plucked and touched are :

A.

Plucked at $\frac{l}{4}$ and touch at $\frac{l}{2}$

B.

Plucked at $\frac{l}{4}$ and touch at $\frac{3 l}{4}$

C.

Plucked at $\frac{l}{2}$ and touched at $\frac{l}{4}$

D.

Plucked at $\frac{l}{2}$ and touched at $\frac{3 l}{4}$

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The tension of a stretched string is increased by 69%. In order to keep its frequency of vibration constant, its length must be increased by :

A.

20%

B.

30%

C.

$\sqrt{69} %$

D.

69%

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Explanation

$n \propto \frac{\sqrt{T}}{l} \Rightarrow l \propto \sqrt{T}$ (As n = constant)

⇒ $\frac{l_{2}}{l_{1}} = \sqrt{\frac{T_{2}}{T_{1}}} = l_{1} \sqrt{\frac{169}{100}} \Rightarrow l_{2} = 1 .3 l_{1} = l_{1} + 30 %$ of l₁

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The length of a sonometer wire tuned to a frequency of 250 Hz is 0.60 metre. The frequency of tuning fork with which the vibrating wire will be in tune when the length is made 0.40 metre is :

A.

250 Hz

B.

375 Hz

C.

256 Hz

D.

384 Hz

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Explanation

$n_{1} l_{1} = n_{2} l_{2} \Rightarrow 250 \times 0 .6 = n_{2} \times 0 .4 \Rightarrow n_{2} = 375$

$\Rightarrow n_{2} = 375 H z$

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Two uniform strings A and B made of steel are made to vibrate under the same tension. If the first overtone of A is equal to the second overtone of B and if the radius of A is twice that of B, the ratio of the lengths of the strings is -

A.

1: 2

B.

1 : 3

C.

1 : 4

D.

1 : 6

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Explanation

First overtone of string A = Second overtone of string B.

⇒ Second harmonic of A = Third harmonic of B

⇒ $n_{2} = n_{3} \Rightarrow {[2 (n_{1})]}_{A} = {[3 (n_{1})]}_{B}$ ( $∴ n_{1} = \frac{1}{2 l} \sqrt{\frac{T}{π r^{2} ρ}}$ )

⇒ $2 [\frac{1}{2 l_{A} r_{A}} \sqrt{\frac{T}{π ρ}}] = 3 [\frac{1}{2 l_{B} r_{B}} \sqrt{\frac{T}{π ρ}}]$

$\frac{l_{A}}{l_{B}} = \frac{2}{3} \frac{r_{B}}{r_{A}} \Rightarrow \frac{l_{A}}{l_{B}} = \frac{2}{3} \times \frac{r_{B}}{(2 r_{B})} = \frac{1}{3}$

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Two wires are fixed in a sonometer. Their tensions are in the ratio 8 : 1. The lengths are in the ratio 36 : 35. The diameters are in the ratio 4 : 1. Densities of the materials are in the ratio 1 : 2. If the lower frequency in the setting is 360 Hz. the beat frequency when the two wires are sounded together is :

A.

5

B.

8

C.

6

D.

10

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Explanation

Frequency in a stretched string is given by $n = \frac{1}{2 l} \sqrt{\frac{T}{π r^{2} ρ}} = \frac{1}{l} \sqrt{\frac{T}{π d^{2} ρ}}$ (d = Diameter of string)

⇒ $\frac{n_{1}}{n_{2}} = \frac{l_{2}}{l_{1}} \sqrt{\frac{T_{1}}{T_{2}} \times {(\frac{d_{2}}{d_{1}})}^{2} \times (\frac{ρ_{2}}{ρ_{1}})}$

$= \frac{35}{36} \sqrt{\frac{8}{1} \times {(\frac{1}{4})}^{2} \times \frac{2}{1}} = \frac{35}{36}$

$\Rightarrow n_{2} = \frac{36}{35} \times 360 = 370$

Hence beat frequency = $n_{2} - n_{1} = 10$

Physics MCQs for NEET — Practice Questions with Answers

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