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JEE Main 2021 (Online) 31st August Morning Shift

Numerical
A wire having a linear mass density 9.0 $$\times$$ 10$$-$$4 kg/m is stretched between two rigid supports with a tension of 900 N. The wire resonates at a frequency of 500 Hz. The next higher frequency at which the same wire resonates is 550 Hz. The length of the wire is ____________ m.

Explanation

$$\mu = 9.0 \times {10^{ - 4}}{{kg} \over m}$$

T = 900 N

$$V = \sqrt {{T \over \mu }} = \sqrt {{{900} \over {9 \times {{10}^{ - 4}}}}} = 1000$$ m/s

f1 = 500 Hz

f = 550

$${{nV} \over {2l}} = 500$$ .... (i)

$${{(n + 1)V} \over {2l}} = 500$$ .... (ii)

(ii) (i) $${V \over {2l}} = 50$$

$$l = {{1000} \over {2 \times 50}} = 10$$
2

JEE Main 2021 (Online) 27th August Evening Shift

Numerical
A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be __________ cm. (Take speed of sound in air as 340 ms$$-$$1)

Explanation

$${\lambda \over 4}$$ = l $$\Rightarrow$$ $$\lambda$$ = 4l

f = $${V \over \lambda } = {V \over {4l}}$$

$$\Rightarrow$$ 250 = $${{340} \over {4l}}$$

$$\Rightarrow$$ l = $${{34} \over {4 \times 25}}$$ = 0.34 m

l = 34 cm
3

JEE Main 2021 (Online) 27th August Morning Shift

Numerical
Two cars X and Y are approaching each other with velocities 36 km/h and 72 km/h respectively. The frequency of a whistle sound as emitted by a passenger in car X, heard by the passenger in car Y is 1320 Hz. If the velocity of sound in air is 340 m/s, the actual frequency of the whistle sound produced is .................. Hz.

Explanation

Image

Vx = 36 km/hr = 10 m/s

Vy = 72 km/hr = 20 m/s

by doppler's effect

$$F' = {F_0}\left( {{{V \pm {V_0}} \over {V \pm {V_s}}}} \right)$$

$$1320 = {F_0}\left( {{{340 + 20} \over {340 - 10}}} \right) \Rightarrow {F_0} = 1210$$ Hz
4

JEE Main 2021 (Online) 26th August Evening Shift

Numerical
Two waves are simultaneously passing through a string and their equations are :

y1 = A1 sin k(x $$-$$ vt), y2 = A2 sin k(x $$-$$ vt + x0). Given amplitudes A1 = 12 mm and A2 = 5 mm, x0 = 3.5 cm and wave number k = 6.28 cm$$-$$1. The amplitude of resulting wave will be ................ mm.

Explanation

y1 = A1 sin k(x $$-$$ vt)

y1 = 12 sin 6.28 (x $$-$$ vt)

y2 = 5 sin 6.28 (x $$-$$ vt + 3.5)

$$\Delta \phi = {{2\pi } \over \lambda }(\Delta x)$$

$$= K(\Delta x)$$

$$= 6.28 \times 3.5 = {7 \over 2} \times 2\pi = 7\pi$$

$${A_{net}} = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi }$$

$${A_{net}} = \sqrt {{{(12)}^2} + {{(5)}^2} + 2(12)(5)\cos (7\pi )}$$

$$= \sqrt {144 + 25 - 120}$$

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