B Given, Young's modulus $=$ Y, Bulk modulus $=\mathrm{K}$, Modulus of elasticity $=\mathrm{q}$ We know that, $Y=3 K(1-2 v)$ $v=\frac{1}{2}\left(1-\frac{Y}{3 K}\right)$ Also, $\quad Y=2 q(1+v)$ $v=\left(\frac{\mathrm{Y}}{2 \mathrm{q}}-1\right)$ Solving equation (i) and (ii), we get - $\frac{1}{2}\left(1-\frac{Y}{3 K}\right)=\left(\frac{Y}{2 q}-1\right)$ $1-\frac{Y}{3 K}=\frac{Y}{q}-2$ $\frac{3}{Y}=\frac{1}{q}+\frac{1}{3 K}$
Shift-I]
Mechanical Properties of Solids
140988
Two different wires made with same material have their radii in the ratio $1: 2$. Even their lengths are in the ratio $1: 2$. On subjecting to different loads, if the extensions produced are equal, find the ratio of the loads applied?
1 $1: 4$
2 $1: 3$
3 $1: 2$
4 $4: 1$
Explanation:
C Given, $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{1}{2}, \frac{l_{1}}{l_{2}}=\frac{1}{2}$ Wires are made by same material $\mathrm{Y}_{1}=\mathrm{Y}_{2}$ $\frac{\mathrm{Y}_{1}}{\mathrm{Y}_{2}}=\frac{\frac{\mathrm{F}_{1} \times l_{1}}{\mathrm{~A}_{1} \times \Delta l_{1}}}{\frac{\mathrm{F}_{2} \times l_{2}}{\mathrm{~A}_{2} \times \Delta l_{2}}}$ $\frac{\mathrm{Y}_{1}}{\mathrm{Y}_{2}}=\frac{\mathrm{F}_{1} \times l_{1} \times \mathrm{A}_{2} \times \Delta l_{2}}{\mathrm{~A}_{1} \times \Delta l_{1} \times \mathrm{F}_{2} \times l_{2}} \quad\left(\because \Delta l_{1}=\Delta l_{2}\right)$ $1=\frac{\mathrm{F}_{1} \times l_{1} \times \mathrm{A}_{2}}{\mathrm{~F}_{2} \times \mathrm{A}_{1} \times l_{2}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{A}_{1} \times l_{2}}{l_{1} \times \mathrm{A}_{2}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\pi \mathrm{r}_{1}^{2} \times l_{2}}{l_{1} \times \pi \mathrm{r}_{2}^{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2} \times \frac{l_{2}}{l_{1}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\left(\frac{1}{2}\right)^{2} \times 2=\frac{1}{2}$ Hence, the ratio of load applied is $1: 2$.
Shift-I]
Mechanical Properties of Solids
140990
If the diameter of a brass rod is $4 \mathrm{~mm}$ and Young's modulus of brass is $9 \times 10^{10} \mathrm{Nm}^{-2}$. Find the force required to stretch it by $0.1 \%$ of its length.
1 $360 \pi \mathrm{N}$
2 $36 \mathrm{~N}$
3 $144 \pi \times 10^{3} \mathrm{~N}$
4 $36 \pi \times 10^{5} \mathrm{~N}$
Explanation:
A Given, diameter of a brass $=4 \mathrm{~mm}$, $\mathrm{Y}=9 \times 10^{10} \mathrm{Nm}^{-2}, \Delta l=0.1 \%$ of $l=\frac{0.1 l}{100}, \mathrm{r}=2 \mathrm{~mm}$ We know that, $\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y} \frac{\Delta l}{l}$ $\mathrm{~F}=\mathrm{AY} \frac{\Delta l}{l}$ $\mathrm{~F}=\pi\left(2 \times 10^{-3}\right)^{2} \times 9 \times 10^{9} \times \frac{1}{100}$ $\mathrm{~F}=\pi \times 4 \times 9 \times 10^{-6} \times 10^{7}$ $\mathrm{~F}=360 \pi \mathrm{N}$
Shift-II]
Mechanical Properties of Solids
140991
What is the pressure required to reduce the given volume of water by $1 \%$ ? $\text { (Bulk modulus }(\mathrm{KB})=\mathbf{2} \times 10^{8} \mathrm{Nm}^{-2} \text { ) }$
B Given, Young's modulus $=$ Y, Bulk modulus $=\mathrm{K}$, Modulus of elasticity $=\mathrm{q}$ We know that, $Y=3 K(1-2 v)$ $v=\frac{1}{2}\left(1-\frac{Y}{3 K}\right)$ Also, $\quad Y=2 q(1+v)$ $v=\left(\frac{\mathrm{Y}}{2 \mathrm{q}}-1\right)$ Solving equation (i) and (ii), we get - $\frac{1}{2}\left(1-\frac{Y}{3 K}\right)=\left(\frac{Y}{2 q}-1\right)$ $1-\frac{Y}{3 K}=\frac{Y}{q}-2$ $\frac{3}{Y}=\frac{1}{q}+\frac{1}{3 K}$
Shift-I]
Mechanical Properties of Solids
140988
Two different wires made with same material have their radii in the ratio $1: 2$. Even their lengths are in the ratio $1: 2$. On subjecting to different loads, if the extensions produced are equal, find the ratio of the loads applied?
1 $1: 4$
2 $1: 3$
3 $1: 2$
4 $4: 1$
Explanation:
C Given, $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{1}{2}, \frac{l_{1}}{l_{2}}=\frac{1}{2}$ Wires are made by same material $\mathrm{Y}_{1}=\mathrm{Y}_{2}$ $\frac{\mathrm{Y}_{1}}{\mathrm{Y}_{2}}=\frac{\frac{\mathrm{F}_{1} \times l_{1}}{\mathrm{~A}_{1} \times \Delta l_{1}}}{\frac{\mathrm{F}_{2} \times l_{2}}{\mathrm{~A}_{2} \times \Delta l_{2}}}$ $\frac{\mathrm{Y}_{1}}{\mathrm{Y}_{2}}=\frac{\mathrm{F}_{1} \times l_{1} \times \mathrm{A}_{2} \times \Delta l_{2}}{\mathrm{~A}_{1} \times \Delta l_{1} \times \mathrm{F}_{2} \times l_{2}} \quad\left(\because \Delta l_{1}=\Delta l_{2}\right)$ $1=\frac{\mathrm{F}_{1} \times l_{1} \times \mathrm{A}_{2}}{\mathrm{~F}_{2} \times \mathrm{A}_{1} \times l_{2}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{A}_{1} \times l_{2}}{l_{1} \times \mathrm{A}_{2}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\pi \mathrm{r}_{1}^{2} \times l_{2}}{l_{1} \times \pi \mathrm{r}_{2}^{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2} \times \frac{l_{2}}{l_{1}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\left(\frac{1}{2}\right)^{2} \times 2=\frac{1}{2}$ Hence, the ratio of load applied is $1: 2$.
Shift-I]
Mechanical Properties of Solids
140990
If the diameter of a brass rod is $4 \mathrm{~mm}$ and Young's modulus of brass is $9 \times 10^{10} \mathrm{Nm}^{-2}$. Find the force required to stretch it by $0.1 \%$ of its length.
1 $360 \pi \mathrm{N}$
2 $36 \mathrm{~N}$
3 $144 \pi \times 10^{3} \mathrm{~N}$
4 $36 \pi \times 10^{5} \mathrm{~N}$
Explanation:
A Given, diameter of a brass $=4 \mathrm{~mm}$, $\mathrm{Y}=9 \times 10^{10} \mathrm{Nm}^{-2}, \Delta l=0.1 \%$ of $l=\frac{0.1 l}{100}, \mathrm{r}=2 \mathrm{~mm}$ We know that, $\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y} \frac{\Delta l}{l}$ $\mathrm{~F}=\mathrm{AY} \frac{\Delta l}{l}$ $\mathrm{~F}=\pi\left(2 \times 10^{-3}\right)^{2} \times 9 \times 10^{9} \times \frac{1}{100}$ $\mathrm{~F}=\pi \times 4 \times 9 \times 10^{-6} \times 10^{7}$ $\mathrm{~F}=360 \pi \mathrm{N}$
Shift-II]
Mechanical Properties of Solids
140991
What is the pressure required to reduce the given volume of water by $1 \%$ ? $\text { (Bulk modulus }(\mathrm{KB})=\mathbf{2} \times 10^{8} \mathrm{Nm}^{-2} \text { ) }$
B Given, Young's modulus $=$ Y, Bulk modulus $=\mathrm{K}$, Modulus of elasticity $=\mathrm{q}$ We know that, $Y=3 K(1-2 v)$ $v=\frac{1}{2}\left(1-\frac{Y}{3 K}\right)$ Also, $\quad Y=2 q(1+v)$ $v=\left(\frac{\mathrm{Y}}{2 \mathrm{q}}-1\right)$ Solving equation (i) and (ii), we get - $\frac{1}{2}\left(1-\frac{Y}{3 K}\right)=\left(\frac{Y}{2 q}-1\right)$ $1-\frac{Y}{3 K}=\frac{Y}{q}-2$ $\frac{3}{Y}=\frac{1}{q}+\frac{1}{3 K}$
Shift-I]
Mechanical Properties of Solids
140988
Two different wires made with same material have their radii in the ratio $1: 2$. Even their lengths are in the ratio $1: 2$. On subjecting to different loads, if the extensions produced are equal, find the ratio of the loads applied?
1 $1: 4$
2 $1: 3$
3 $1: 2$
4 $4: 1$
Explanation:
C Given, $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{1}{2}, \frac{l_{1}}{l_{2}}=\frac{1}{2}$ Wires are made by same material $\mathrm{Y}_{1}=\mathrm{Y}_{2}$ $\frac{\mathrm{Y}_{1}}{\mathrm{Y}_{2}}=\frac{\frac{\mathrm{F}_{1} \times l_{1}}{\mathrm{~A}_{1} \times \Delta l_{1}}}{\frac{\mathrm{F}_{2} \times l_{2}}{\mathrm{~A}_{2} \times \Delta l_{2}}}$ $\frac{\mathrm{Y}_{1}}{\mathrm{Y}_{2}}=\frac{\mathrm{F}_{1} \times l_{1} \times \mathrm{A}_{2} \times \Delta l_{2}}{\mathrm{~A}_{1} \times \Delta l_{1} \times \mathrm{F}_{2} \times l_{2}} \quad\left(\because \Delta l_{1}=\Delta l_{2}\right)$ $1=\frac{\mathrm{F}_{1} \times l_{1} \times \mathrm{A}_{2}}{\mathrm{~F}_{2} \times \mathrm{A}_{1} \times l_{2}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{A}_{1} \times l_{2}}{l_{1} \times \mathrm{A}_{2}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\pi \mathrm{r}_{1}^{2} \times l_{2}}{l_{1} \times \pi \mathrm{r}_{2}^{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2} \times \frac{l_{2}}{l_{1}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\left(\frac{1}{2}\right)^{2} \times 2=\frac{1}{2}$ Hence, the ratio of load applied is $1: 2$.
Shift-I]
Mechanical Properties of Solids
140990
If the diameter of a brass rod is $4 \mathrm{~mm}$ and Young's modulus of brass is $9 \times 10^{10} \mathrm{Nm}^{-2}$. Find the force required to stretch it by $0.1 \%$ of its length.
1 $360 \pi \mathrm{N}$
2 $36 \mathrm{~N}$
3 $144 \pi \times 10^{3} \mathrm{~N}$
4 $36 \pi \times 10^{5} \mathrm{~N}$
Explanation:
A Given, diameter of a brass $=4 \mathrm{~mm}$, $\mathrm{Y}=9 \times 10^{10} \mathrm{Nm}^{-2}, \Delta l=0.1 \%$ of $l=\frac{0.1 l}{100}, \mathrm{r}=2 \mathrm{~mm}$ We know that, $\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y} \frac{\Delta l}{l}$ $\mathrm{~F}=\mathrm{AY} \frac{\Delta l}{l}$ $\mathrm{~F}=\pi\left(2 \times 10^{-3}\right)^{2} \times 9 \times 10^{9} \times \frac{1}{100}$ $\mathrm{~F}=\pi \times 4 \times 9 \times 10^{-6} \times 10^{7}$ $\mathrm{~F}=360 \pi \mathrm{N}$
Shift-II]
Mechanical Properties of Solids
140991
What is the pressure required to reduce the given volume of water by $1 \%$ ? $\text { (Bulk modulus }(\mathrm{KB})=\mathbf{2} \times 10^{8} \mathrm{Nm}^{-2} \text { ) }$
B Given, Young's modulus $=$ Y, Bulk modulus $=\mathrm{K}$, Modulus of elasticity $=\mathrm{q}$ We know that, $Y=3 K(1-2 v)$ $v=\frac{1}{2}\left(1-\frac{Y}{3 K}\right)$ Also, $\quad Y=2 q(1+v)$ $v=\left(\frac{\mathrm{Y}}{2 \mathrm{q}}-1\right)$ Solving equation (i) and (ii), we get - $\frac{1}{2}\left(1-\frac{Y}{3 K}\right)=\left(\frac{Y}{2 q}-1\right)$ $1-\frac{Y}{3 K}=\frac{Y}{q}-2$ $\frac{3}{Y}=\frac{1}{q}+\frac{1}{3 K}$
Shift-I]
Mechanical Properties of Solids
140988
Two different wires made with same material have their radii in the ratio $1: 2$. Even their lengths are in the ratio $1: 2$. On subjecting to different loads, if the extensions produced are equal, find the ratio of the loads applied?
1 $1: 4$
2 $1: 3$
3 $1: 2$
4 $4: 1$
Explanation:
C Given, $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{1}{2}, \frac{l_{1}}{l_{2}}=\frac{1}{2}$ Wires are made by same material $\mathrm{Y}_{1}=\mathrm{Y}_{2}$ $\frac{\mathrm{Y}_{1}}{\mathrm{Y}_{2}}=\frac{\frac{\mathrm{F}_{1} \times l_{1}}{\mathrm{~A}_{1} \times \Delta l_{1}}}{\frac{\mathrm{F}_{2} \times l_{2}}{\mathrm{~A}_{2} \times \Delta l_{2}}}$ $\frac{\mathrm{Y}_{1}}{\mathrm{Y}_{2}}=\frac{\mathrm{F}_{1} \times l_{1} \times \mathrm{A}_{2} \times \Delta l_{2}}{\mathrm{~A}_{1} \times \Delta l_{1} \times \mathrm{F}_{2} \times l_{2}} \quad\left(\because \Delta l_{1}=\Delta l_{2}\right)$ $1=\frac{\mathrm{F}_{1} \times l_{1} \times \mathrm{A}_{2}}{\mathrm{~F}_{2} \times \mathrm{A}_{1} \times l_{2}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{A}_{1} \times l_{2}}{l_{1} \times \mathrm{A}_{2}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\pi \mathrm{r}_{1}^{2} \times l_{2}}{l_{1} \times \pi \mathrm{r}_{2}^{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2} \times \frac{l_{2}}{l_{1}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\left(\frac{1}{2}\right)^{2} \times 2=\frac{1}{2}$ Hence, the ratio of load applied is $1: 2$.
Shift-I]
Mechanical Properties of Solids
140990
If the diameter of a brass rod is $4 \mathrm{~mm}$ and Young's modulus of brass is $9 \times 10^{10} \mathrm{Nm}^{-2}$. Find the force required to stretch it by $0.1 \%$ of its length.
1 $360 \pi \mathrm{N}$
2 $36 \mathrm{~N}$
3 $144 \pi \times 10^{3} \mathrm{~N}$
4 $36 \pi \times 10^{5} \mathrm{~N}$
Explanation:
A Given, diameter of a brass $=4 \mathrm{~mm}$, $\mathrm{Y}=9 \times 10^{10} \mathrm{Nm}^{-2}, \Delta l=0.1 \%$ of $l=\frac{0.1 l}{100}, \mathrm{r}=2 \mathrm{~mm}$ We know that, $\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y} \frac{\Delta l}{l}$ $\mathrm{~F}=\mathrm{AY} \frac{\Delta l}{l}$ $\mathrm{~F}=\pi\left(2 \times 10^{-3}\right)^{2} \times 9 \times 10^{9} \times \frac{1}{100}$ $\mathrm{~F}=\pi \times 4 \times 9 \times 10^{-6} \times 10^{7}$ $\mathrm{~F}=360 \pi \mathrm{N}$
Shift-II]
Mechanical Properties of Solids
140991
What is the pressure required to reduce the given volume of water by $1 \%$ ? $\text { (Bulk modulus }(\mathrm{KB})=\mathbf{2} \times 10^{8} \mathrm{Nm}^{-2} \text { ) }$
B Given, Young's modulus $=$ Y, Bulk modulus $=\mathrm{K}$, Modulus of elasticity $=\mathrm{q}$ We know that, $Y=3 K(1-2 v)$ $v=\frac{1}{2}\left(1-\frac{Y}{3 K}\right)$ Also, $\quad Y=2 q(1+v)$ $v=\left(\frac{\mathrm{Y}}{2 \mathrm{q}}-1\right)$ Solving equation (i) and (ii), we get - $\frac{1}{2}\left(1-\frac{Y}{3 K}\right)=\left(\frac{Y}{2 q}-1\right)$ $1-\frac{Y}{3 K}=\frac{Y}{q}-2$ $\frac{3}{Y}=\frac{1}{q}+\frac{1}{3 K}$
Shift-I]
Mechanical Properties of Solids
140988
Two different wires made with same material have their radii in the ratio $1: 2$. Even their lengths are in the ratio $1: 2$. On subjecting to different loads, if the extensions produced are equal, find the ratio of the loads applied?
1 $1: 4$
2 $1: 3$
3 $1: 2$
4 $4: 1$
Explanation:
C Given, $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{1}{2}, \frac{l_{1}}{l_{2}}=\frac{1}{2}$ Wires are made by same material $\mathrm{Y}_{1}=\mathrm{Y}_{2}$ $\frac{\mathrm{Y}_{1}}{\mathrm{Y}_{2}}=\frac{\frac{\mathrm{F}_{1} \times l_{1}}{\mathrm{~A}_{1} \times \Delta l_{1}}}{\frac{\mathrm{F}_{2} \times l_{2}}{\mathrm{~A}_{2} \times \Delta l_{2}}}$ $\frac{\mathrm{Y}_{1}}{\mathrm{Y}_{2}}=\frac{\mathrm{F}_{1} \times l_{1} \times \mathrm{A}_{2} \times \Delta l_{2}}{\mathrm{~A}_{1} \times \Delta l_{1} \times \mathrm{F}_{2} \times l_{2}} \quad\left(\because \Delta l_{1}=\Delta l_{2}\right)$ $1=\frac{\mathrm{F}_{1} \times l_{1} \times \mathrm{A}_{2}}{\mathrm{~F}_{2} \times \mathrm{A}_{1} \times l_{2}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{A}_{1} \times l_{2}}{l_{1} \times \mathrm{A}_{2}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\pi \mathrm{r}_{1}^{2} \times l_{2}}{l_{1} \times \pi \mathrm{r}_{2}^{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2} \times \frac{l_{2}}{l_{1}}$ $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\left(\frac{1}{2}\right)^{2} \times 2=\frac{1}{2}$ Hence, the ratio of load applied is $1: 2$.
Shift-I]
Mechanical Properties of Solids
140990
If the diameter of a brass rod is $4 \mathrm{~mm}$ and Young's modulus of brass is $9 \times 10^{10} \mathrm{Nm}^{-2}$. Find the force required to stretch it by $0.1 \%$ of its length.
1 $360 \pi \mathrm{N}$
2 $36 \mathrm{~N}$
3 $144 \pi \times 10^{3} \mathrm{~N}$
4 $36 \pi \times 10^{5} \mathrm{~N}$
Explanation:
A Given, diameter of a brass $=4 \mathrm{~mm}$, $\mathrm{Y}=9 \times 10^{10} \mathrm{Nm}^{-2}, \Delta l=0.1 \%$ of $l=\frac{0.1 l}{100}, \mathrm{r}=2 \mathrm{~mm}$ We know that, $\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y} \frac{\Delta l}{l}$ $\mathrm{~F}=\mathrm{AY} \frac{\Delta l}{l}$ $\mathrm{~F}=\pi\left(2 \times 10^{-3}\right)^{2} \times 9 \times 10^{9} \times \frac{1}{100}$ $\mathrm{~F}=\pi \times 4 \times 9 \times 10^{-6} \times 10^{7}$ $\mathrm{~F}=360 \pi \mathrm{N}$
Shift-II]
Mechanical Properties of Solids
140991
What is the pressure required to reduce the given volume of water by $1 \%$ ? $\text { (Bulk modulus }(\mathrm{KB})=\mathbf{2} \times 10^{8} \mathrm{Nm}^{-2} \text { ) }$
