148800
The efficiency of a heat engine is $1 / 6$. Its when the temperature of sink decreases by $62^{\circ} \mathrm{C}$, its efficiency doubles. Then what is the temperature of sources?
1 $470 \mathrm{~K}$
2 $372 \mathrm{~K}$
3 $542 \mathrm{~K}$
4 $1042 \mathrm{~K}$
Explanation:
B $\eta=\frac{1}{6}, \eta^{\prime}=2 \eta$ Let $T_{1}=$ Temperature of sink Given, $\mathrm{T}_{2}=$ Temperature of source $\mathrm{T}_{2}^{\prime}=\mathrm{T}_{2}-62$ We know that, The efficiency of the Carnot engine can be given as, $\eta =1-\frac{\mathrm{T}_{2}(\mathrm{~K})}{\mathrm{T}_{1}(\mathrm{~K})}$ For $=\frac{1}{6},$ $\frac{1}{6} =1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} =\frac{5}{6}$ For $\quad \eta^{\prime}=2 \eta$, $\eta^{\prime} =1-\frac{T_{2}^{\prime}}{T_{1}}$ $2 \eta =1-\frac{T_{2}-62}{T_{1}}$ $\frac{1}{3} =1-\frac{T_{2}}{T_{1}}+\frac{62}{T_{1}}$ $\frac{1}{3} =1-\frac{5}{6}+\frac{62}{T_{1}}$ $\frac{62}{T_{1}} =\frac{1}{3}-\frac{1}{6}=\frac{1}{6}$ $T_{1} =6 \times 62$ $=372 \mathrm{~K}$
JIPMER-2018
Work, Energy and Power
148801
A Carnot engine absorbs $6 \times 10^{5}$ cal at $227^{\circ} \mathrm{C}$. The work done per cycle by the engine, if its sink is maintained at $127^{\circ} \mathrm{C}$ is
148802
When a rubber string is stretched through a distance ' $x$ ', the restoring force developed has a magnitude $\left(p x+q x^2+r x^3\right)$ where $p, q$ and $r$ are constants. Work done in stretching the unstretched rubber string by a distance ' $l$ ' is
A Given, $\mathrm{F}=\mathrm{px}+\mathrm{qx}^{2}+\mathrm{rx}^{3}$ We know that, Work done $(\mathrm{dW})=\mathrm{F} \mathrm{dx}$ $\int_{0}^{\mathrm{w}} \mathrm{dW}=\int_{0}^{l}\left(\mathrm{px}+\mathrm{qx}^{2}+\mathrm{rx}^{3}\right) \mathrm{dx}$ $\mathrm{W}=\left[\frac{\mathrm{p} x^{2}}{2}+\frac{\mathrm{q} x^{3}}{3}+\frac{\mathrm{r} x^{4}}{4}\right]_{0}^{l}$ $\mathrm{W}=\frac{\mathrm{p} l^{2}}{2}+\frac{\mathrm{q} l^{3}}{3}+\frac{\mathrm{r} l^{4}}{4}$
AP EAMCET-25.04.2018
Work, Energy and Power
148803
A body of mass $2 \mathrm{~kg}$ moving in $\mathrm{X}-\mathrm{Y}$ plane has a potential energy given by $U=(6 x+8 y) J$. The body is at rest at the point $(3,2) \mathrm{m}$. The work to be done by the body to reach another position after $2 \mathrm{~s}$ is
1 $100 \mathrm{~J}$
2 $500 \mathrm{~J}$
3 $750 \mathrm{~J}$
4 $900 \mathrm{~J}$
Explanation:
A Given, Potential energy, $\mathrm{U}=(6 \mathrm{x}+8 \mathrm{y}) \mathrm{J}$ Mass $(\mathrm{m})=2 \mathrm{~kg}$. Time $(\mathrm{t})=2 \mathrm{~s}$. Body is rest at point $=(3,2)$ $\because \mathrm{U}=6 \mathrm{x}+8 \mathrm{y}$. Now, $F=\frac{-\partial U}{\partial x} \hat{i}+\frac{-\partial U}{\partial y} \hat{j}$ $F=-6 \hat{i}-8 \hat{j}$ $\therefore a=\frac{F}{m}=\frac{-6 \hat{i}-8 \hat{j}}{2}=-3 \hat{i}-4 \hat{j}$ We know that. $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$. $\therefore \mathrm{x}=\mathrm{x}_{0}+\frac{1}{2} \mathrm{a}_{\mathrm{x}} \mathrm{t}^{2}=3+\frac{1}{2}(-3)(2)^{2}=-3 \mathrm{~m} \quad\left[\mathrm{x}_{\mathrm{o}}=3\right]$ Similarly, $\mathrm{y}=\mathrm{y}_{0}+\frac{1}{2} \mathrm{a}_{\mathrm{y}} \mathrm{t}^{2}=2+\frac{1}{2}(-4)(2)^{2}=-6 \mathrm{~m} \quad\left[\mathrm{y}_{\mathrm{o}}=3\right]$ $\therefore \mathrm{U}_{0}=6 \mathrm{x}_{0}+8 \mathrm{y}_{0}=6 \times 3+8 \times 2=18+16=34 \mathrm{~J}$ $\mathrm{U}_{1}=6 \mathrm{x}+8 \mathrm{y}=6 \times-3+8 \times-6=-18-48=-66 \mathrm{~J}$ $\therefore \Delta \mathrm{U}=\mathrm{U}_{0}-\mathrm{U}_{1}=34-(-66)=100 \mathrm{~J}$
148800
The efficiency of a heat engine is $1 / 6$. Its when the temperature of sink decreases by $62^{\circ} \mathrm{C}$, its efficiency doubles. Then what is the temperature of sources?
1 $470 \mathrm{~K}$
2 $372 \mathrm{~K}$
3 $542 \mathrm{~K}$
4 $1042 \mathrm{~K}$
Explanation:
B $\eta=\frac{1}{6}, \eta^{\prime}=2 \eta$ Let $T_{1}=$ Temperature of sink Given, $\mathrm{T}_{2}=$ Temperature of source $\mathrm{T}_{2}^{\prime}=\mathrm{T}_{2}-62$ We know that, The efficiency of the Carnot engine can be given as, $\eta =1-\frac{\mathrm{T}_{2}(\mathrm{~K})}{\mathrm{T}_{1}(\mathrm{~K})}$ For $=\frac{1}{6},$ $\frac{1}{6} =1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} =\frac{5}{6}$ For $\quad \eta^{\prime}=2 \eta$, $\eta^{\prime} =1-\frac{T_{2}^{\prime}}{T_{1}}$ $2 \eta =1-\frac{T_{2}-62}{T_{1}}$ $\frac{1}{3} =1-\frac{T_{2}}{T_{1}}+\frac{62}{T_{1}}$ $\frac{1}{3} =1-\frac{5}{6}+\frac{62}{T_{1}}$ $\frac{62}{T_{1}} =\frac{1}{3}-\frac{1}{6}=\frac{1}{6}$ $T_{1} =6 \times 62$ $=372 \mathrm{~K}$
JIPMER-2018
Work, Energy and Power
148801
A Carnot engine absorbs $6 \times 10^{5}$ cal at $227^{\circ} \mathrm{C}$. The work done per cycle by the engine, if its sink is maintained at $127^{\circ} \mathrm{C}$ is
148802
When a rubber string is stretched through a distance ' $x$ ', the restoring force developed has a magnitude $\left(p x+q x^2+r x^3\right)$ where $p, q$ and $r$ are constants. Work done in stretching the unstretched rubber string by a distance ' $l$ ' is
A Given, $\mathrm{F}=\mathrm{px}+\mathrm{qx}^{2}+\mathrm{rx}^{3}$ We know that, Work done $(\mathrm{dW})=\mathrm{F} \mathrm{dx}$ $\int_{0}^{\mathrm{w}} \mathrm{dW}=\int_{0}^{l}\left(\mathrm{px}+\mathrm{qx}^{2}+\mathrm{rx}^{3}\right) \mathrm{dx}$ $\mathrm{W}=\left[\frac{\mathrm{p} x^{2}}{2}+\frac{\mathrm{q} x^{3}}{3}+\frac{\mathrm{r} x^{4}}{4}\right]_{0}^{l}$ $\mathrm{W}=\frac{\mathrm{p} l^{2}}{2}+\frac{\mathrm{q} l^{3}}{3}+\frac{\mathrm{r} l^{4}}{4}$
AP EAMCET-25.04.2018
Work, Energy and Power
148803
A body of mass $2 \mathrm{~kg}$ moving in $\mathrm{X}-\mathrm{Y}$ plane has a potential energy given by $U=(6 x+8 y) J$. The body is at rest at the point $(3,2) \mathrm{m}$. The work to be done by the body to reach another position after $2 \mathrm{~s}$ is
1 $100 \mathrm{~J}$
2 $500 \mathrm{~J}$
3 $750 \mathrm{~J}$
4 $900 \mathrm{~J}$
Explanation:
A Given, Potential energy, $\mathrm{U}=(6 \mathrm{x}+8 \mathrm{y}) \mathrm{J}$ Mass $(\mathrm{m})=2 \mathrm{~kg}$. Time $(\mathrm{t})=2 \mathrm{~s}$. Body is rest at point $=(3,2)$ $\because \mathrm{U}=6 \mathrm{x}+8 \mathrm{y}$. Now, $F=\frac{-\partial U}{\partial x} \hat{i}+\frac{-\partial U}{\partial y} \hat{j}$ $F=-6 \hat{i}-8 \hat{j}$ $\therefore a=\frac{F}{m}=\frac{-6 \hat{i}-8 \hat{j}}{2}=-3 \hat{i}-4 \hat{j}$ We know that. $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$. $\therefore \mathrm{x}=\mathrm{x}_{0}+\frac{1}{2} \mathrm{a}_{\mathrm{x}} \mathrm{t}^{2}=3+\frac{1}{2}(-3)(2)^{2}=-3 \mathrm{~m} \quad\left[\mathrm{x}_{\mathrm{o}}=3\right]$ Similarly, $\mathrm{y}=\mathrm{y}_{0}+\frac{1}{2} \mathrm{a}_{\mathrm{y}} \mathrm{t}^{2}=2+\frac{1}{2}(-4)(2)^{2}=-6 \mathrm{~m} \quad\left[\mathrm{y}_{\mathrm{o}}=3\right]$ $\therefore \mathrm{U}_{0}=6 \mathrm{x}_{0}+8 \mathrm{y}_{0}=6 \times 3+8 \times 2=18+16=34 \mathrm{~J}$ $\mathrm{U}_{1}=6 \mathrm{x}+8 \mathrm{y}=6 \times-3+8 \times-6=-18-48=-66 \mathrm{~J}$ $\therefore \Delta \mathrm{U}=\mathrm{U}_{0}-\mathrm{U}_{1}=34-(-66)=100 \mathrm{~J}$
148800
The efficiency of a heat engine is $1 / 6$. Its when the temperature of sink decreases by $62^{\circ} \mathrm{C}$, its efficiency doubles. Then what is the temperature of sources?
1 $470 \mathrm{~K}$
2 $372 \mathrm{~K}$
3 $542 \mathrm{~K}$
4 $1042 \mathrm{~K}$
Explanation:
B $\eta=\frac{1}{6}, \eta^{\prime}=2 \eta$ Let $T_{1}=$ Temperature of sink Given, $\mathrm{T}_{2}=$ Temperature of source $\mathrm{T}_{2}^{\prime}=\mathrm{T}_{2}-62$ We know that, The efficiency of the Carnot engine can be given as, $\eta =1-\frac{\mathrm{T}_{2}(\mathrm{~K})}{\mathrm{T}_{1}(\mathrm{~K})}$ For $=\frac{1}{6},$ $\frac{1}{6} =1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} =\frac{5}{6}$ For $\quad \eta^{\prime}=2 \eta$, $\eta^{\prime} =1-\frac{T_{2}^{\prime}}{T_{1}}$ $2 \eta =1-\frac{T_{2}-62}{T_{1}}$ $\frac{1}{3} =1-\frac{T_{2}}{T_{1}}+\frac{62}{T_{1}}$ $\frac{1}{3} =1-\frac{5}{6}+\frac{62}{T_{1}}$ $\frac{62}{T_{1}} =\frac{1}{3}-\frac{1}{6}=\frac{1}{6}$ $T_{1} =6 \times 62$ $=372 \mathrm{~K}$
JIPMER-2018
Work, Energy and Power
148801
A Carnot engine absorbs $6 \times 10^{5}$ cal at $227^{\circ} \mathrm{C}$. The work done per cycle by the engine, if its sink is maintained at $127^{\circ} \mathrm{C}$ is
148802
When a rubber string is stretched through a distance ' $x$ ', the restoring force developed has a magnitude $\left(p x+q x^2+r x^3\right)$ where $p, q$ and $r$ are constants. Work done in stretching the unstretched rubber string by a distance ' $l$ ' is
A Given, $\mathrm{F}=\mathrm{px}+\mathrm{qx}^{2}+\mathrm{rx}^{3}$ We know that, Work done $(\mathrm{dW})=\mathrm{F} \mathrm{dx}$ $\int_{0}^{\mathrm{w}} \mathrm{dW}=\int_{0}^{l}\left(\mathrm{px}+\mathrm{qx}^{2}+\mathrm{rx}^{3}\right) \mathrm{dx}$ $\mathrm{W}=\left[\frac{\mathrm{p} x^{2}}{2}+\frac{\mathrm{q} x^{3}}{3}+\frac{\mathrm{r} x^{4}}{4}\right]_{0}^{l}$ $\mathrm{W}=\frac{\mathrm{p} l^{2}}{2}+\frac{\mathrm{q} l^{3}}{3}+\frac{\mathrm{r} l^{4}}{4}$
AP EAMCET-25.04.2018
Work, Energy and Power
148803
A body of mass $2 \mathrm{~kg}$ moving in $\mathrm{X}-\mathrm{Y}$ plane has a potential energy given by $U=(6 x+8 y) J$. The body is at rest at the point $(3,2) \mathrm{m}$. The work to be done by the body to reach another position after $2 \mathrm{~s}$ is
1 $100 \mathrm{~J}$
2 $500 \mathrm{~J}$
3 $750 \mathrm{~J}$
4 $900 \mathrm{~J}$
Explanation:
A Given, Potential energy, $\mathrm{U}=(6 \mathrm{x}+8 \mathrm{y}) \mathrm{J}$ Mass $(\mathrm{m})=2 \mathrm{~kg}$. Time $(\mathrm{t})=2 \mathrm{~s}$. Body is rest at point $=(3,2)$ $\because \mathrm{U}=6 \mathrm{x}+8 \mathrm{y}$. Now, $F=\frac{-\partial U}{\partial x} \hat{i}+\frac{-\partial U}{\partial y} \hat{j}$ $F=-6 \hat{i}-8 \hat{j}$ $\therefore a=\frac{F}{m}=\frac{-6 \hat{i}-8 \hat{j}}{2}=-3 \hat{i}-4 \hat{j}$ We know that. $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$. $\therefore \mathrm{x}=\mathrm{x}_{0}+\frac{1}{2} \mathrm{a}_{\mathrm{x}} \mathrm{t}^{2}=3+\frac{1}{2}(-3)(2)^{2}=-3 \mathrm{~m} \quad\left[\mathrm{x}_{\mathrm{o}}=3\right]$ Similarly, $\mathrm{y}=\mathrm{y}_{0}+\frac{1}{2} \mathrm{a}_{\mathrm{y}} \mathrm{t}^{2}=2+\frac{1}{2}(-4)(2)^{2}=-6 \mathrm{~m} \quad\left[\mathrm{y}_{\mathrm{o}}=3\right]$ $\therefore \mathrm{U}_{0}=6 \mathrm{x}_{0}+8 \mathrm{y}_{0}=6 \times 3+8 \times 2=18+16=34 \mathrm{~J}$ $\mathrm{U}_{1}=6 \mathrm{x}+8 \mathrm{y}=6 \times-3+8 \times-6=-18-48=-66 \mathrm{~J}$ $\therefore \Delta \mathrm{U}=\mathrm{U}_{0}-\mathrm{U}_{1}=34-(-66)=100 \mathrm{~J}$
148800
The efficiency of a heat engine is $1 / 6$. Its when the temperature of sink decreases by $62^{\circ} \mathrm{C}$, its efficiency doubles. Then what is the temperature of sources?
1 $470 \mathrm{~K}$
2 $372 \mathrm{~K}$
3 $542 \mathrm{~K}$
4 $1042 \mathrm{~K}$
Explanation:
B $\eta=\frac{1}{6}, \eta^{\prime}=2 \eta$ Let $T_{1}=$ Temperature of sink Given, $\mathrm{T}_{2}=$ Temperature of source $\mathrm{T}_{2}^{\prime}=\mathrm{T}_{2}-62$ We know that, The efficiency of the Carnot engine can be given as, $\eta =1-\frac{\mathrm{T}_{2}(\mathrm{~K})}{\mathrm{T}_{1}(\mathrm{~K})}$ For $=\frac{1}{6},$ $\frac{1}{6} =1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} =\frac{5}{6}$ For $\quad \eta^{\prime}=2 \eta$, $\eta^{\prime} =1-\frac{T_{2}^{\prime}}{T_{1}}$ $2 \eta =1-\frac{T_{2}-62}{T_{1}}$ $\frac{1}{3} =1-\frac{T_{2}}{T_{1}}+\frac{62}{T_{1}}$ $\frac{1}{3} =1-\frac{5}{6}+\frac{62}{T_{1}}$ $\frac{62}{T_{1}} =\frac{1}{3}-\frac{1}{6}=\frac{1}{6}$ $T_{1} =6 \times 62$ $=372 \mathrm{~K}$
JIPMER-2018
Work, Energy and Power
148801
A Carnot engine absorbs $6 \times 10^{5}$ cal at $227^{\circ} \mathrm{C}$. The work done per cycle by the engine, if its sink is maintained at $127^{\circ} \mathrm{C}$ is
148802
When a rubber string is stretched through a distance ' $x$ ', the restoring force developed has a magnitude $\left(p x+q x^2+r x^3\right)$ where $p, q$ and $r$ are constants. Work done in stretching the unstretched rubber string by a distance ' $l$ ' is
A Given, $\mathrm{F}=\mathrm{px}+\mathrm{qx}^{2}+\mathrm{rx}^{3}$ We know that, Work done $(\mathrm{dW})=\mathrm{F} \mathrm{dx}$ $\int_{0}^{\mathrm{w}} \mathrm{dW}=\int_{0}^{l}\left(\mathrm{px}+\mathrm{qx}^{2}+\mathrm{rx}^{3}\right) \mathrm{dx}$ $\mathrm{W}=\left[\frac{\mathrm{p} x^{2}}{2}+\frac{\mathrm{q} x^{3}}{3}+\frac{\mathrm{r} x^{4}}{4}\right]_{0}^{l}$ $\mathrm{W}=\frac{\mathrm{p} l^{2}}{2}+\frac{\mathrm{q} l^{3}}{3}+\frac{\mathrm{r} l^{4}}{4}$
AP EAMCET-25.04.2018
Work, Energy and Power
148803
A body of mass $2 \mathrm{~kg}$ moving in $\mathrm{X}-\mathrm{Y}$ plane has a potential energy given by $U=(6 x+8 y) J$. The body is at rest at the point $(3,2) \mathrm{m}$. The work to be done by the body to reach another position after $2 \mathrm{~s}$ is
1 $100 \mathrm{~J}$
2 $500 \mathrm{~J}$
3 $750 \mathrm{~J}$
4 $900 \mathrm{~J}$
Explanation:
A Given, Potential energy, $\mathrm{U}=(6 \mathrm{x}+8 \mathrm{y}) \mathrm{J}$ Mass $(\mathrm{m})=2 \mathrm{~kg}$. Time $(\mathrm{t})=2 \mathrm{~s}$. Body is rest at point $=(3,2)$ $\because \mathrm{U}=6 \mathrm{x}+8 \mathrm{y}$. Now, $F=\frac{-\partial U}{\partial x} \hat{i}+\frac{-\partial U}{\partial y} \hat{j}$ $F=-6 \hat{i}-8 \hat{j}$ $\therefore a=\frac{F}{m}=\frac{-6 \hat{i}-8 \hat{j}}{2}=-3 \hat{i}-4 \hat{j}$ We know that. $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$. $\therefore \mathrm{x}=\mathrm{x}_{0}+\frac{1}{2} \mathrm{a}_{\mathrm{x}} \mathrm{t}^{2}=3+\frac{1}{2}(-3)(2)^{2}=-3 \mathrm{~m} \quad\left[\mathrm{x}_{\mathrm{o}}=3\right]$ Similarly, $\mathrm{y}=\mathrm{y}_{0}+\frac{1}{2} \mathrm{a}_{\mathrm{y}} \mathrm{t}^{2}=2+\frac{1}{2}(-4)(2)^{2}=-6 \mathrm{~m} \quad\left[\mathrm{y}_{\mathrm{o}}=3\right]$ $\therefore \mathrm{U}_{0}=6 \mathrm{x}_{0}+8 \mathrm{y}_{0}=6 \times 3+8 \times 2=18+16=34 \mathrm{~J}$ $\mathrm{U}_{1}=6 \mathrm{x}+8 \mathrm{y}=6 \times-3+8 \times-6=-18-48=-66 \mathrm{~J}$ $\therefore \Delta \mathrm{U}=\mathrm{U}_{0}-\mathrm{U}_{1}=34-(-66)=100 \mathrm{~J}$
