JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Elasticity & SHM

The density and breaking stress of a wire are

6 \times 1 0^{4} kg / m^{3}

and

1.2 \times 1 0^{8} N / m^{2}

respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity is

\frac{1}{3}

of the value on the surface of earth. The maximum length of the wire with breaking is ______ m (take, g =

10 m / s^{2}

)

Answer: 600

Solution

The wire breaks under its own weight when the stress at the top equals the breaking stress:

σ = \frac{( A L ρ ) g ^{'}}{A} = ρ g^{'} L

. With

g^{'} = \frac{g}{3} = \frac{10}{3} m s^{- 2}

L_{ma x} = \frac{σ}{ρ g ^{'}} = \frac{1.2 \times 1 0 ^{8}}{( 6 \times 1 0 ^{4} ) ( 10/3 )} = \frac{1.2 \times 1 0 ^{8}}{2 \times 1 0 ^{5}} = 600 m .

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Integer answer