JEE MAIN - Physics (2021 - 31st August Morning Shift - No. 22)
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.
Answer
10
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) $${V \over {2l}} = 50$$
$$l = {{1000} \over {2 \times 50}} = 10$$
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) $${V \over {2l}} = 50$$
$$l = {{1000} \over {2 \times 50}} = 10$$
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