JEE Main 2025 Apr 2 Shift 2, Physics Q16: Newton's Laws Of Motion

JEE Main2025Apr 2, Shift 2Physics

A body of mass $1$ kg is suspended with the help of two strings making angles as shown in figure. Magnitude of tensions $T_{1}$ and $T_{2}$ , respectively, are (in N) :

A
$5, 53$
B
$53, 5$
C
$53, 53$
D
$5, 5$

Solution

Because the two strings make $6 0^{\circ}$ and $3 0^{\circ}$ with the ceiling, they are mutually perpendicular at the block. The weight $m g$ then resolves cleanly along the two string directions.

Component of weight along $T_{1}$ (string at $6 0^{\circ}$ to ceiling, i.e. $3 0^{\circ}$ to the vertical): $T_{1} = m g cos 3 0^{\circ}$ .

Component of weight along $T_{2}$ (string at $3 0^{\circ}$ to ceiling, i.e. $6 0^{\circ}$ to the vertical): $T_{2} = m g sin 3 0^{\circ}$ ... equivalently $m g cos 6 0^{\circ}$ .

With $m = 1$ kg and $g = 10$ m/s $^{2}$ (so $m g = 10$ N):

$T_{1} = 10 \cdot \frac{3}{2} = 53$ N,

$T_{2} = 10 \cdot \frac{1}{2} = 5$ N.

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JEE Main 2025 Apr 2 Shift 2, Physics Q16: Newton's Laws Of Motion

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