JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 5)
In an experiment to determine the period of a simple pendulum of length 1 m, it is attached to different spherical bobs of radii r1
and r2 . The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be
5×10−4 s, the difference in radii, $$\left| {} \right.$$r1 $$-$$ r2 $$\left| {} \right.$$ is best given by :
1 cm
0.05 cm
0.5 cm
0.01 cm
Explanation
The time period is given by the formula
$$T = 2\pi \sqrt {{l \over g}} $$
which clearly indicates that the time period is directly proportional to the length of the pendulum l, that is,
$$T \propto \sqrt l $$
Here, l = 1 m. Therefore,
$${{\Delta T} \over l} = {1 \over 2}{{\Delta l} \over l}$$ ....... (1)
where $$\Delta l = \left| {{r_1} - {r_2}} \right|$$. Substituting the values in Eq. (1), we get
$$5 \times {10^{ - 4}} = {1 \over 2}\left( {{{{r_1} - {r_2}} \over 1}} \right)$$
$$ \Rightarrow {r_1} - {r_2} = 10 \times {10^{ - 4}} = {10^{ - 3}}$$ m $$ = {10^{ - 1}}$$ cm $$ = 0.1$$ cm
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