JEE PHYSICS MCQ: THERMAL PROPERTIES OF MATTER

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Application +4 | -1

What force should be applied to the ends of steel rod of a cross sectional area 10 cm2 to revent it from elongation when heated from 273K to 303K.

\left( {\alpha = 1 \times {{10}^{ - 5}}/{\,^{\,0}}C,y = 2 \times {{10}^{11}}\,w/{m^2}} \right)

View answer and explanation

Correct option: C

6 \times {10^4}N

Option C (

6 \times {10^4}N

) is correct because it is the force required to prevent elongation, calculated using the formula

F = yA\alpha \Delta t

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