Question

A uniform metre scale of mass 700 g is suspended horizontally using two strings tied at 20 cm and 70 cm marks on the scale. The tension in the string at 70 cm mark is (Acceleration due to gravity = 10 m/s$^2$):

• A. 2.8 N
• B. 280 N
• C. 4.2 N
• D. 420 N

Hint:

For equilibrium, sum of clockwise moments equals sum of anticlockwise moments. Always take moments about one support to eliminate its reaction. The center of gravity of a uniform rod lies at its midpoint. Check for consistent distance units (cm or m) before calculating.

Answer 1

The Correct Option is C

• Let the tensions at 20 cm and 70 cm be $T_1$ and $T_2$ respectively.
• The weight of the metre scale, $W = mg = 0.7 \times 10 = 7$ N, acts at the center (50 cm mark).
• Taking moments about the 20 cm mark: \[ T_2 \times (70 - 20) = 7 \times (50 - 20) \] \[ T_2 \times 50 = 7 \times 30 \Rightarrow T_2 = \frac{210}{50} = 4.2 \text{ N} \] • Hence, the tension at the 70 cm mark is 4.2 N.