JEE Main 2021 (Online) 25th February Morning Shift Physics - Circular Motion Question 47 English Explanation

Let the speed of bob at lowest position be v₁ and at the highest position be v₂.

Maximum tension is at lowest position and minimum tension is at the highest position. Now, using, conservation of mechanical energy,

$${1 \over 2}mv_1^2 = {1 \over 2}mv_2^2 + mg2l$$

$$ \Rightarrow {v_1}^2 = {v_2}^2 + 4gl$$ ..........(1)

Now, $${T_{\max }} - mg = {{mv_1^2} \over l}$$

$$ \Rightarrow {T_{\max }} = mg + {{mv_1^2} \over l}$$

& $${T_{\min }} + mg = {{mv_2^2} \over l}$$

$$ \Rightarrow {T_{\min }} = {{mv_2^2} \over l} - mg$$

$${{{T_{\max }}} \over {{T_{\min }}}} = {5 \over 1}$$

$$ \Rightarrow {{mg + {{mv_1^2} \over l}} \over {{{mv_2^2} \over l} - mg}} = {5 \over 1}$$

$$ \Rightarrow mg + {{mv_1^2} \over l} = \left[ {{{mv_2^2} \over l} - mg} \right]5$$

$$ \Rightarrow mg + {m \over l}\left[ {v_2^2 + 4gl} \right] = {{5mv_2^2} \over l} - 5mg$$

$$ \Rightarrow mg + {{mv_2^2} \over l} + 4mg = {{5mv_2^2} \over l} - 5mg$$

$$ \Rightarrow 10mg = {{4mv_2^2} \over l}$$

$${v_2}^2 = {{10 \times 10 \times 1} \over 4}$$

$$ \Rightarrow {v_2}^2 = 25 \Rightarrow {v_2} = 5$$ m/s

Thus, velocity of bob at highest position 5 m/s.