A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass ' m ' is suspended from the rod at 160 cm mark as shown in the figure Find the value of 'm' such that the rod is in equilibrium. ( g =10 m / s 2)

Question

A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass ' m ' is suspended from the rod at 160 cm mark as shown in the figure Find the value of 'm' such that the rod is in equilibrium. ( g =10 m / s 2) (A) (1/2) kg (B) (1/3) kg (C) (1/6) kg (D) (1/12) kg

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a1e8ab9b5adac0fad1cca826410ad268-.png" /> From principle of moments 2 × 20=0.5 × 60+ m × 120 2=1.5+6 m 0.5=6 m m =(1/12) kg

$\frac{1}{2} k g$

$\frac{1}{3} k g$

$\frac{1}{6} k g$

$\frac{1}{12} k g$

From principle of moments
$2 \times 20 = 0.5 \times 60 + m \times 120$
$2 = 1.5 + 6 m$
$0.5 = 6 m$
$m = \frac{1}{12} k g$

21​kg

31​kg

61​kg

121​kg

From principle of moments 2×20=0.5×60+m×120 2=1.5+6m 0.5=6m m=121​kg

$\frac{1}{2} k g$

$\frac{1}{3} k g$

$\frac{1}{6} k g$

$\frac{1}{12} k g$

From principle of moments
$2 \times 20 = 0.5 \times 60 + m \times 120$
$2 = 1.5 + 6 m$
$0.5 = 6 m$
$m = \frac{1}{12} k g$