AP EAMCET · PHYSICS · Mechanical Properties of Solids

The force required to stretch a steel wire of area of cross-section $1 \mathrm{~mm}^2$ to double its length is
(Young's modulus of steel $=2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$ )

A $2 \times 10^3 \mathrm{~N}$
B $2 \times 10^5 \mathrm{~N}$
C $2 \times 10^2 \mathrm{~N}$
D $2 \times 10^4 \mathrm{~N}$

Verified Solution

Answer & Solution

Correct Answer

(B) $2 \times 10^5 \mathrm{~N}$

Step-by-step Solution

Detailed explanation

$F = Y A \frac{\Delta L}{L}$ $F = (2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}) \times (1 \times 10^{-6} \mathrm{~m}^2) \times (1)$ $F = 2 \times 10^5 \mathrm{~N}$

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match the following?

	Column $- I$		Column $- II$
(a)	Ratio of change in time period of a simple pendulum with temperature to its original time period	(i)	$α Δ T$
(b)	Ratio of the value of a length to its scale reading	(ii)	$T$
(c)	Reciprocal of coefficient of volume expansion for an ideal gas of constant pressure	(iii)	$(1 + α Δ T)$
(d)	$\frac{F}{Y A} =$	(iv)	$\frac{1}{2} α Δ T$

AP EAMCET 2021 Medium

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The force required to stretch a steel wire of area of cross-section \(1 \mathrm{~mm}^2\) to double its length is
(Young's modulus of steel \(=2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\) )

Answer & Solution

Step-by-step Solution

See the Complete Solution

The force required to stretch a steel wire of area of cross-section \(1 \mathrm{~mm}^2\) to double its length is(Young's modulus of steel \(=2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\) )

Step-by-step Solution

The force required to stretch a steel wire of area of cross-section \(1 \mathrm{~mm}^2\) to double its length is
(Young's modulus of steel \(=2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\) )