Question

The length of the scale of a spring balance that can weigh zero to 100 kg is 25 cm. If a body suspended from this balance oscillates with a time period of $0.2\pi$ s, then the mass of the suspended body is (Acceleration due to gravity = $10 \text{ m s}^{-2}$)

• A. 30 kg
• B. 50 kg
• C. 40 kg
• D. 60 kg

Hint:

For problems involving springs and oscillations, remember two key relationships: Hooke's Law ($F=kx$) to find the spring constant, and the time period formula for a mass-spring system ($T=2\pi\sqrt{m/k}$). Ensure all units are consistent (SI units) throughout the calculation.

Answer 1

The Correct Option is C

Step 1: Determine the spring constant (k) of the spring balance.
The spring balance can weigh up to 100 kg over a scale length of 25 cm. This means that when a mass of 100 kg is suspended, the spring extends by 25 cm due to the gravitational force acting on it. Force exerted by 100 kg mass ($F$) = $m_{max} \times g$ $F = 100 \text{ kg} \times 10 \text{ m s}^{-2} = 1000 \text{ N}$. The extension ($x$) for this force is $25 \text{ cm} = 0.25 \text{ m}$. According to Hooke's Law, the force exerted by a spring is $F = kx$, where $k$ is the spring constant. \[ k = \frac{F}{x} = \frac{1000 \text{ N}}{0.25 \text{ m}} \] \[ k = 4000 \text{ N/m} \] Step 2: Use the formula for the time period of oscillation of a mass-spring system.
When a body of mass 'm' is suspended from the spring and oscillates, its time period ($T$) is given by: \[ T = 2\pi \sqrt{\frac{m}{k}} \] Step 3: Calculate the mass (m) of the suspended body.
We are given the time period $T = 0.2\pi \text{ s}$ and we have calculated $k = 4000 \text{ N/m}$. Substitute these values into the time period formula: \[ 0.2\pi = 2\pi \sqrt{\frac{m}{4000}} \] Divide both sides by $2\pi$: \[ 0.1 = \sqrt{\frac{m}{4000}} \] Square both sides to remove the square root: \[ (0.1)^2 = \frac{m}{4000} \] \[ 0.01 = \frac{m}{4000} \] Now, solve for $m$: \[ m = 0.01 \times 4000 \] \[ m = 40 \text{ kg} \] The final answer is $\boxed{40 \text{ kg}}$.