JEE Advance - Physics (2020 - Paper 1 Offline - No. 3)
A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with
a speed of 2 ms−1
in front of the open end of the pipe and parallel to it, the length of the pipe should
be changed for the resonance to occur with the moving tuning fork. If the speed of sound in air is
320 ms−1, the smallest value of the percentage change required in the length of the pipe is
____________.
Одговорити
0.62to0.63
Објашњење
$$f \propto {1 \over {{l_1}}} \Rightarrow f = {k \over {{l_1}}}$$ ....(i)
(where, l1 $$ \Rightarrow $$ initial length of pipe)
$$\left( {{v \over {v - {v_T}}}} \right)f = {k \over {{l_1}}}$$ ....(ii)
(where, vT = speed of tuning fork, l2 = new length of pipe)
Dividing Eq. (i) by Eq. (ii), we get
$${{v - {v_T}} \over c} = {{{l_2}} \over {{l_1}}} \Rightarrow {{{l_2}} \over {{l_1}}} - 1 = {{v - {v_T}} \over v} - 1$$
$${{{l_2} - {l_1}} \over {{l_1}}} = {{ - {v_T}} \over v}$$
$${{{l_2} - {l_1}} \over {{l_1}}} \times 100 = {{ - 2} \over {320}} \times 100 = - 0.625$$
Therefore, smallest value of percentage change required in the length of pipe is 0.625.
(where, l1 $$ \Rightarrow $$ initial length of pipe)
$$\left( {{v \over {v - {v_T}}}} \right)f = {k \over {{l_1}}}$$ ....(ii)
(where, vT = speed of tuning fork, l2 = new length of pipe)
Dividing Eq. (i) by Eq. (ii), we get
$${{v - {v_T}} \over c} = {{{l_2}} \over {{l_1}}} \Rightarrow {{{l_2}} \over {{l_1}}} - 1 = {{v - {v_T}} \over v} - 1$$
$${{{l_2} - {l_1}} \over {{l_1}}} = {{ - {v_T}} \over v}$$
$${{{l_2} - {l_1}} \over {{l_1}}} \times 100 = {{ - 2} \over {320}} \times 100 = - 0.625$$
Therefore, smallest value of percentage change required in the length of pipe is 0.625.