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### 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 $$\,\,\, k = mf2 \times 4\pi 2 Given, f = 1012 m = {{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}} \therefore\,\,\, k = {{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}} \times 1012 \times 4\pi 2 = 7.1 N/m 4 MCQ (Single Correct Answer) ### JEE Main 2018 (Offline) A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 \times 103 kg/m3 and its Young’s modulus is 9.27 \times 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ? A 7.5 kHz B 5 kHz C 2.5 kHz D 10 kHz ## Explanation As rod length = 60 cm \therefore\,\,\, {\lambda \over 2} = 60 \Rightarrow$$\,\,\,$ $\lambda$ = 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