Question

The figure shows a rectangular plate of dimension $(l\times b)$ from which two circular holes of radii $R_1$ and $R_2(R_1>R_2)$ has been cut. The distance between the two holes is $d(<R_2).$ Now the rod is heated up. If $l^{\prime },$ $b^{\prime },R_1^{\prime },R_2^{\prime }$ and $d^{\prime }$ are the respective lengths at a higher temperature. Then after heating

	Column-I		Column-II
p	$\frac{l^{\prime }}{l}$	a	1
q	$\frac{R_1^{\prime }}{R_1}$	b	$(1+\alpha \Delta T)$
r	$\frac{R_2^{\prime }}{R_2}$	c	$(1-\alpha \Delta T)$
s	$\frac{d^{\prime }}{d}$	d	$\left(\alpha \Delta T\right)$

• A. p-a ; q-c ; r-c ; s-b
• B. p-c ; q-c ; r-a ; s-d
• C. p-b ; q-a ; r-a ; s-b
• D. p-b ; q-b ; r-b ; s-b

Answer 1

The correct option is D p-b ; q-b ; r-b ; s-b

Thermal expansion of an isotropic object may be imagined as a photographic enlargement. Hence all the dimensions increase as the temperature is increased.
Let the temperature be increased by $\Delta T$
$$l^{\prime }=l(1+\alpha \Delta T)\Rightarrow \frac{l^{\prime }}{l}=(1+\alpha \Delta T)$$ $b^{\prime }=b(1+\alpha \Delta T)\Rightarrow \frac{b^{\prime }}{b}=(1+\alpha \Delta T)$
$R_1^{\prime }=R_1(1+\alpha \Delta T)\Rightarrow \frac{R_1^{\prime }}{R_1}=(1+\alpha \Delta T)$
$R_2^{\prime }=R_2(1+\alpha \Delta T)\Rightarrow \frac{R_2^{\prime }}{R_2}=(1+\alpha \Delta T)$
$d^{\prime }=d(1+\alpha \Delta T)\Rightarrow \frac{d^{\prime }}{d}=(1+\alpha \Delta T)$
Hence, all these ratios can be represented as
$\frac{l^{\prime }}{l}=\frac{b^{\prime }}{b}=\frac{R_2^{\prime }}{R_1}=\frac{R_2^{\prime }}{R_2}=\frac{d^{\prime }}{d}=1+\alpha \Delta T$