1
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 8th April Morning Slot

Two wires W1 and W2 have the same radius r and respective densities $$\rho $$1 and $$\rho $$2 such that ρ2 = 4$$\rho $$1 . They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in W1 to W2 is :

A
1 : 1
B
1 : 2
C
1 : 3
D
4 : 1

Explanation

Fundamental frequency of a stretched string is ,

    f = $${n \over {2L}}\sqrt {{T \over \mu }} $$

Here, n = number of antinodes

$$\mu $$ = mass per unit length.

As, O is the midpoint of two bridegs, hence length of two wires are equal.

$$\therefore\,\,\,$$ L1 = L2 = L

As frequency of both wires same

f1 = f2

$$ \Rightarrow $$$$\,\,\,$$ $${{{n_1}} \over {2L}}\sqrt {{T \over {\pi {r^2}{\rho _1}}}} $$ = $${{{n_2}} \over {2L}}\sqrt {{T \over {\pi {r^2}{\rho _2}}}} $$

$$ \Rightarrow $$$$\,\,\,$$ $${{{n_1}} \over {{n_2}}}$$ = $$\sqrt {{{{\rho _1}} \over {{\rho _2}}}} $$

$$ \Rightarrow $$$$\,\,\,$$ $${{{n_1}} \over {{n_2}}}$$ = $$\sqrt {{{{\rho _1}} \over {4{\rho _1}}}} $$ = $${1 \over 2}$$
2
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 9th April Morning Slot

A standing wave is formed by the superposition of two waves travelling in opposite directions. The transverse displacement is given by

y(x, t) = 0.5 sin $$\left( {{{5\pi } \over 4}x} \right)\,$$ cos(200 $$\pi $$t).

What is the speed of the travelling wave moving in the positive x direction ?

(x and t are in meter and second, respectively.)
A
160 m/s
B
90 m/s
C
180 m/s
D
120 m/s

Explanation

Standard equation of standing wave,

y(x, t) = 2a sin kx cos $$\omega $$t

Given,

y(x, t) = 0.5 sin $$\left( {{{5\pi } \over 4}x} \right)$$ cos (200$$\pi $$t).

So, k = $${{5\pi } \over 4}$$ and $$\omega $$ = 200$$\pi $$

$$\therefore\,\,\,$$ Speed of travelling wave

= $${\omega \over k}$$ = $${{200\pi } \over {{{5\pi } \over 4}}}$$ = 160 m/s.
3
MCQ (Single Correct Answer)

JEE Main 2018 (Offline)

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avogadro number = 6.02 × 1023 gm mole–1)
A
5.5 N/m
B
6.4 N/m
C
7.1 N/m
D
2.2 N/m

Explanation

6.02 $$ \times $$ 1023 atoms of silver = 108 gm

1   atoms   of   silver   =   $${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$$ kg

For a harmonic oscillator

f = $${1 \over {2\pi }}$$ $$\sqrt {{k \over m}} $$

Where  k = force constant

$$ \Rightarrow $$$$\,\,\,$$ f2 = $${1 \over {4{\pi ^2}}}$$ $$\left( {{k \over m}} \right)$$

$$ \Rightarrow $$$$\,\,\,$$ k = mf2 $$ \times $$ 4$$\pi $$2

Given,

f = 1012

m = $${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$$

$$\therefore\,\,\,$$ k = $${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$$ $$ \times $$ 1012 $$ \times $$ 4$$\pi $$2

= 7.1 N/m
4
MCQ (Single Correct Answer)

JEE Main 2018 (Offline)

A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 $$\times$$ 103 kg/m3 and its Young’s modulus is 9.27 $$\times$$ 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?
A
7.5 kHz
B
5 kHz
C
2.5 kHz
D
10 kHz

Explanation



As   rod length = 60 cm

$$\therefore\,\,\,$$ $${\lambda \over 2}$$ = 60

$$ \Rightarrow $$$$\,\,\,$$ $$\lambda $$ = 120 cm = 1.2 m

In solid, velocity of wave,

V = $$\sqrt {{Y \over \rho }} $$

= $$\sqrt {{{9.27 \times {{10}^{10}}} \over {2.7 \times {{10}^3}}}} $$

= 5.85 $$ \times $$ 103 m/sec.

As  we know,

v = f $$\lambda $$

$$\therefore\,\,\,$$ f = $${v \over \lambda }$$

= $${{5.85 \times {{10}^3}} \over {1.2}}$$

= 4.88 $$ \times $$ 103 Hz

$$ \simeq $$  5 kHz

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