Question

A metal block of mass m is suspended from a rigid support through a metal wire of diameter 14 mm. The tensile stress developed in the wire under equilibrium state is 7× 10⁵ Nm^-2 . The value of mass m is ___ kg. (Take, g = 9.8 ms-2 and π =\(\frac{ 22}{7}\))

Answer 1

Correct Answer: 11

To find the mass \(m\) of the metal block given the tensile stress in the wire, we follow these steps:

1. **Tensile Stress Formula**: The tensile stress (\(\sigma\)) developed in the wire is given by the formula:   \[\sigma = \frac{F}{A}\] where \(F\) is the force due to the weight of the block, and \(A\) is the cross-sectional area of the wire.

2. **Determine the Force**: The force \(F\) is due to the weight of the block, hence \(F = mg\), where \(g = 9.8\, \text{m/s}^2\) is the acceleration due to gravity.

3. **Calculate Cross-Sectional Area**: The cross-sectional area \(A\) of the wire with diameter \(d = 14\, \text{mm} = 0.014\, \text{m}\) is given by:  \[A = \frac{\pi d^2}{4} = \frac{22}{7} \times \left(\frac{0.014}{2}\right)^2 = \frac{22}{7} \times 0.007^2 = \frac{22}{7} \times 0.000049 = 0.000154\, \text{m}^2\]

4. **Calculate the Mass**: Using the tensile stress formula \(\sigma = \frac{mg}{A}\), rearrange to find \(m\): \[m = \frac{\sigma \times A}{g} = \frac{7 \times 10^5 \times 0.000154}{9.8} = \frac{107.8}{9.8} \approx 11\, \text{kg}\]

5. **Verification**: The calculated mass \(11\, \text{kg}\) is within the expected range of [11, 11].

Therefore, the mass of the metal block is \(11\, \text{kg}\).