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Physics - Elasticity Question with Solution | TestHub

PhysicsElasticityMiscellaneousEasy2 minPYQ_2023

PhysicsEasynumerical

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 $7 \times 10^{5} N m^{- 2}$ . The value of mass $m$ is ______ $kg$ .

(Take $g = 9.8 m s^{- 2}$ and $π = \frac{22}{7}$ )

Answer:

11.00

Solution:

The data given is
$T = 7 \times 10^{5} N m^{- 2} r = 7 \times 10^{- 3} m$
Using the equation
$\frac{m g}{A} = T \Rightarrow m g = T A \Rightarrow m g = T π r^{2}$
$\Rightarrow m g = 7 \times 10^{5} \times \frac{22}{7} \times 7^{2} \times 10^{- 6}$
$\Rightarrow m g = \frac{49 \times 22}{10}$
$\Rightarrow m = \frac{49 \times 22}{98} = 11 kg$

Stream:JEESubject:PhysicsTopic:ElasticitySubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2023

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