JEE MAIN - Physics (2003 - No. 57)
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads $$49$$ $$N,$$ when the lift is stationary. If the lift moves downward with an acceleration of $$5 m/{s^2}$$, the reading of the spring balance will be
$$24$$ $$N$$
$$74$$ $$N$$
$$15$$ $$N$$
$$49$$ $$N$$
Explanation
When lift is stationary then,
T1 = $$mg$$ = 49 N
$$m$$ = 5
For the bag accelerating down
$$mg-T=ma$$
$$\therefore$$ $$T=m(g-a)$$
$$ = 5\left( {9.8 - 5} \right) = 24\,N$$
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