146201
An engine of power \(58.8 \mathrm{~kW}\) pulls a train of mass \(2 \times 10^{5} \mathrm{~kg}\) with a velocity of \(36 \mathrm{kmh}^{-1}\). The coefficient of friction is
1 0.3
2 0.03
3 0.003
4 0.0003
5 0.04
Explanation:
C Given that, Power of engine \((\mathrm{P})=58.8 \times 10^{3} \mathrm{~W}\) Mass of Train \((\mathrm{m})=2 \times 10^{5} \mathrm{~kg}\) Velocity of Train \((\mathrm{v})=36 \mathrm{~km} / \mathrm{h}=\frac{36 \times 5}{18}=10 \mathrm{~m} / \mathrm{s}\) We know that, Power \((\mathrm{P})=\) F.V \(\therefore \quad \mathrm{F}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{58.8 \times 10^{3}}{10}=5880 \mathrm{~N}\) Let, the coefficient of friction, \(\therefore \quad \mathrm{F}=\mathrm{f}_{\mathrm{S}}=\mu \mathrm{mg}\) \(\mu=\frac{\mathrm{F}}{\mathrm{mg}}=\frac{5880}{2 \times 10^{5} \times 10}=0.00294 \approx 0.003\)
Kerala CEE - 2016
Laws of Motion
146202
A mass of \(1 \mathrm{~kg}\) is just able to slide down the slope of an inclined rough surface when the angle of inclination is \(60^{\circ}\). The minimum force necessary to pull the mass up the inclined plane \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) is
146203
A car of mass \(1000 \mathrm{~kg}\) moves on a circular track of radius \(20 \mathrm{~m}\). If the coefficient of friction is 0.64 , then the maximum velocity with which the car can move is :
1 \(15 \mathrm{~m} / \mathrm{s}\)
2 \(11.2 \mathrm{~m} / \mathrm{s}\)
3 \(20 \mathrm{~m} / \mathrm{s}\)
4 \(18 \mathrm{~m} / \mathrm{s}\)
5 \(22.4 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given that, Coefficient of friction \((\mu)=0.64\) Radius of circular track \((r)=20 \mathrm{~m}\) Gravitational acceleration of the body \((\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}\) Mass \(=1000 \mathrm{~kg}\) According the relation between centripetal force and frictional force Frictional force \(=\) Centripetal force \(\mu \mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) \(\mathrm{v}^{2}=\mu \mathrm{rg}\) \(\qquad\ v =\sqrt{\mu \mathrm{rg}}\) \(\text { From eq }^{\mathrm{n}} (\mathrm{i}), \text { the maximum velocity }\) \(\mathrm{v} =\sqrt{0.64 \times 20 \times 10}\) \(=\sqrt{2 \times 64}=8 \sqrt{2}\) \(=8 \times 1.414=11.31 \mathrm{~m} / \mathrm{s} \approx 11.2 \mathrm{~m} / \mathrm{s}\)
Kerala CEE 2006
Laws of Motion
146204
A uniform metal chain is placed on a rough table such that one end of it hangs down over the edge of the table. When one-third of its length hangs over the edge, the chain starts sliding. Then, the coefficient of static friction is:
1 \(\frac{3}{4}\)
2 \(\frac{1}{4}\)
3 \(\frac{2}{3}\)
4 \(\frac{1}{3}\)
5 \(\frac{1}{2}\)
Explanation:
E Let the mass of the chain be M. When \(\frac{1}{3}\) of the chain hangs down the edges magnitude of the force on the chain in the downward direction is equal to the weight of \(\frac{1}{3}\) of the chain i.e. \(\mathrm{Mg} / 3\) Then, the mass of chain portion of the chain lying the table is \(2 \mathrm{M} / 3\). The magnitude of normal reaction is \(2 \mathrm{Mg} / 3\). \(\therefore\) Maximum force of friction on the chain \(=\mu_{\mathrm{s}}\) \((2 \mathrm{Mg} / 3)\). When the chain just starts sliding Force on the chain hangs down the edges \(=\) Force on the chain lying the table \(\therefore \quad \mathrm{Mg} / 3=\mu_{\mathrm{s}}(2 \mathrm{Mg} / 3)\) \(\mu_{\mathrm{s}}=\frac{1}{2}\)
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Laws of Motion
146201
An engine of power \(58.8 \mathrm{~kW}\) pulls a train of mass \(2 \times 10^{5} \mathrm{~kg}\) with a velocity of \(36 \mathrm{kmh}^{-1}\). The coefficient of friction is
1 0.3
2 0.03
3 0.003
4 0.0003
5 0.04
Explanation:
C Given that, Power of engine \((\mathrm{P})=58.8 \times 10^{3} \mathrm{~W}\) Mass of Train \((\mathrm{m})=2 \times 10^{5} \mathrm{~kg}\) Velocity of Train \((\mathrm{v})=36 \mathrm{~km} / \mathrm{h}=\frac{36 \times 5}{18}=10 \mathrm{~m} / \mathrm{s}\) We know that, Power \((\mathrm{P})=\) F.V \(\therefore \quad \mathrm{F}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{58.8 \times 10^{3}}{10}=5880 \mathrm{~N}\) Let, the coefficient of friction, \(\therefore \quad \mathrm{F}=\mathrm{f}_{\mathrm{S}}=\mu \mathrm{mg}\) \(\mu=\frac{\mathrm{F}}{\mathrm{mg}}=\frac{5880}{2 \times 10^{5} \times 10}=0.00294 \approx 0.003\)
Kerala CEE - 2016
Laws of Motion
146202
A mass of \(1 \mathrm{~kg}\) is just able to slide down the slope of an inclined rough surface when the angle of inclination is \(60^{\circ}\). The minimum force necessary to pull the mass up the inclined plane \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) is
146203
A car of mass \(1000 \mathrm{~kg}\) moves on a circular track of radius \(20 \mathrm{~m}\). If the coefficient of friction is 0.64 , then the maximum velocity with which the car can move is :
1 \(15 \mathrm{~m} / \mathrm{s}\)
2 \(11.2 \mathrm{~m} / \mathrm{s}\)
3 \(20 \mathrm{~m} / \mathrm{s}\)
4 \(18 \mathrm{~m} / \mathrm{s}\)
5 \(22.4 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given that, Coefficient of friction \((\mu)=0.64\) Radius of circular track \((r)=20 \mathrm{~m}\) Gravitational acceleration of the body \((\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}\) Mass \(=1000 \mathrm{~kg}\) According the relation between centripetal force and frictional force Frictional force \(=\) Centripetal force \(\mu \mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) \(\mathrm{v}^{2}=\mu \mathrm{rg}\) \(\qquad\ v =\sqrt{\mu \mathrm{rg}}\) \(\text { From eq }^{\mathrm{n}} (\mathrm{i}), \text { the maximum velocity }\) \(\mathrm{v} =\sqrt{0.64 \times 20 \times 10}\) \(=\sqrt{2 \times 64}=8 \sqrt{2}\) \(=8 \times 1.414=11.31 \mathrm{~m} / \mathrm{s} \approx 11.2 \mathrm{~m} / \mathrm{s}\)
Kerala CEE 2006
Laws of Motion
146204
A uniform metal chain is placed on a rough table such that one end of it hangs down over the edge of the table. When one-third of its length hangs over the edge, the chain starts sliding. Then, the coefficient of static friction is:
1 \(\frac{3}{4}\)
2 \(\frac{1}{4}\)
3 \(\frac{2}{3}\)
4 \(\frac{1}{3}\)
5 \(\frac{1}{2}\)
Explanation:
E Let the mass of the chain be M. When \(\frac{1}{3}\) of the chain hangs down the edges magnitude of the force on the chain in the downward direction is equal to the weight of \(\frac{1}{3}\) of the chain i.e. \(\mathrm{Mg} / 3\) Then, the mass of chain portion of the chain lying the table is \(2 \mathrm{M} / 3\). The magnitude of normal reaction is \(2 \mathrm{Mg} / 3\). \(\therefore\) Maximum force of friction on the chain \(=\mu_{\mathrm{s}}\) \((2 \mathrm{Mg} / 3)\). When the chain just starts sliding Force on the chain hangs down the edges \(=\) Force on the chain lying the table \(\therefore \quad \mathrm{Mg} / 3=\mu_{\mathrm{s}}(2 \mathrm{Mg} / 3)\) \(\mu_{\mathrm{s}}=\frac{1}{2}\)
146201
An engine of power \(58.8 \mathrm{~kW}\) pulls a train of mass \(2 \times 10^{5} \mathrm{~kg}\) with a velocity of \(36 \mathrm{kmh}^{-1}\). The coefficient of friction is
1 0.3
2 0.03
3 0.003
4 0.0003
5 0.04
Explanation:
C Given that, Power of engine \((\mathrm{P})=58.8 \times 10^{3} \mathrm{~W}\) Mass of Train \((\mathrm{m})=2 \times 10^{5} \mathrm{~kg}\) Velocity of Train \((\mathrm{v})=36 \mathrm{~km} / \mathrm{h}=\frac{36 \times 5}{18}=10 \mathrm{~m} / \mathrm{s}\) We know that, Power \((\mathrm{P})=\) F.V \(\therefore \quad \mathrm{F}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{58.8 \times 10^{3}}{10}=5880 \mathrm{~N}\) Let, the coefficient of friction, \(\therefore \quad \mathrm{F}=\mathrm{f}_{\mathrm{S}}=\mu \mathrm{mg}\) \(\mu=\frac{\mathrm{F}}{\mathrm{mg}}=\frac{5880}{2 \times 10^{5} \times 10}=0.00294 \approx 0.003\)
Kerala CEE - 2016
Laws of Motion
146202
A mass of \(1 \mathrm{~kg}\) is just able to slide down the slope of an inclined rough surface when the angle of inclination is \(60^{\circ}\). The minimum force necessary to pull the mass up the inclined plane \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) is
146203
A car of mass \(1000 \mathrm{~kg}\) moves on a circular track of radius \(20 \mathrm{~m}\). If the coefficient of friction is 0.64 , then the maximum velocity with which the car can move is :
1 \(15 \mathrm{~m} / \mathrm{s}\)
2 \(11.2 \mathrm{~m} / \mathrm{s}\)
3 \(20 \mathrm{~m} / \mathrm{s}\)
4 \(18 \mathrm{~m} / \mathrm{s}\)
5 \(22.4 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given that, Coefficient of friction \((\mu)=0.64\) Radius of circular track \((r)=20 \mathrm{~m}\) Gravitational acceleration of the body \((\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}\) Mass \(=1000 \mathrm{~kg}\) According the relation between centripetal force and frictional force Frictional force \(=\) Centripetal force \(\mu \mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) \(\mathrm{v}^{2}=\mu \mathrm{rg}\) \(\qquad\ v =\sqrt{\mu \mathrm{rg}}\) \(\text { From eq }^{\mathrm{n}} (\mathrm{i}), \text { the maximum velocity }\) \(\mathrm{v} =\sqrt{0.64 \times 20 \times 10}\) \(=\sqrt{2 \times 64}=8 \sqrt{2}\) \(=8 \times 1.414=11.31 \mathrm{~m} / \mathrm{s} \approx 11.2 \mathrm{~m} / \mathrm{s}\)
Kerala CEE 2006
Laws of Motion
146204
A uniform metal chain is placed on a rough table such that one end of it hangs down over the edge of the table. When one-third of its length hangs over the edge, the chain starts sliding. Then, the coefficient of static friction is:
1 \(\frac{3}{4}\)
2 \(\frac{1}{4}\)
3 \(\frac{2}{3}\)
4 \(\frac{1}{3}\)
5 \(\frac{1}{2}\)
Explanation:
E Let the mass of the chain be M. When \(\frac{1}{3}\) of the chain hangs down the edges magnitude of the force on the chain in the downward direction is equal to the weight of \(\frac{1}{3}\) of the chain i.e. \(\mathrm{Mg} / 3\) Then, the mass of chain portion of the chain lying the table is \(2 \mathrm{M} / 3\). The magnitude of normal reaction is \(2 \mathrm{Mg} / 3\). \(\therefore\) Maximum force of friction on the chain \(=\mu_{\mathrm{s}}\) \((2 \mathrm{Mg} / 3)\). When the chain just starts sliding Force on the chain hangs down the edges \(=\) Force on the chain lying the table \(\therefore \quad \mathrm{Mg} / 3=\mu_{\mathrm{s}}(2 \mathrm{Mg} / 3)\) \(\mu_{\mathrm{s}}=\frac{1}{2}\)
146201
An engine of power \(58.8 \mathrm{~kW}\) pulls a train of mass \(2 \times 10^{5} \mathrm{~kg}\) with a velocity of \(36 \mathrm{kmh}^{-1}\). The coefficient of friction is
1 0.3
2 0.03
3 0.003
4 0.0003
5 0.04
Explanation:
C Given that, Power of engine \((\mathrm{P})=58.8 \times 10^{3} \mathrm{~W}\) Mass of Train \((\mathrm{m})=2 \times 10^{5} \mathrm{~kg}\) Velocity of Train \((\mathrm{v})=36 \mathrm{~km} / \mathrm{h}=\frac{36 \times 5}{18}=10 \mathrm{~m} / \mathrm{s}\) We know that, Power \((\mathrm{P})=\) F.V \(\therefore \quad \mathrm{F}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{58.8 \times 10^{3}}{10}=5880 \mathrm{~N}\) Let, the coefficient of friction, \(\therefore \quad \mathrm{F}=\mathrm{f}_{\mathrm{S}}=\mu \mathrm{mg}\) \(\mu=\frac{\mathrm{F}}{\mathrm{mg}}=\frac{5880}{2 \times 10^{5} \times 10}=0.00294 \approx 0.003\)
Kerala CEE - 2016
Laws of Motion
146202
A mass of \(1 \mathrm{~kg}\) is just able to slide down the slope of an inclined rough surface when the angle of inclination is \(60^{\circ}\). The minimum force necessary to pull the mass up the inclined plane \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) is
146203
A car of mass \(1000 \mathrm{~kg}\) moves on a circular track of radius \(20 \mathrm{~m}\). If the coefficient of friction is 0.64 , then the maximum velocity with which the car can move is :
1 \(15 \mathrm{~m} / \mathrm{s}\)
2 \(11.2 \mathrm{~m} / \mathrm{s}\)
3 \(20 \mathrm{~m} / \mathrm{s}\)
4 \(18 \mathrm{~m} / \mathrm{s}\)
5 \(22.4 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given that, Coefficient of friction \((\mu)=0.64\) Radius of circular track \((r)=20 \mathrm{~m}\) Gravitational acceleration of the body \((\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}\) Mass \(=1000 \mathrm{~kg}\) According the relation between centripetal force and frictional force Frictional force \(=\) Centripetal force \(\mu \mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) \(\mathrm{v}^{2}=\mu \mathrm{rg}\) \(\qquad\ v =\sqrt{\mu \mathrm{rg}}\) \(\text { From eq }^{\mathrm{n}} (\mathrm{i}), \text { the maximum velocity }\) \(\mathrm{v} =\sqrt{0.64 \times 20 \times 10}\) \(=\sqrt{2 \times 64}=8 \sqrt{2}\) \(=8 \times 1.414=11.31 \mathrm{~m} / \mathrm{s} \approx 11.2 \mathrm{~m} / \mathrm{s}\)
Kerala CEE 2006
Laws of Motion
146204
A uniform metal chain is placed on a rough table such that one end of it hangs down over the edge of the table. When one-third of its length hangs over the edge, the chain starts sliding. Then, the coefficient of static friction is:
1 \(\frac{3}{4}\)
2 \(\frac{1}{4}\)
3 \(\frac{2}{3}\)
4 \(\frac{1}{3}\)
5 \(\frac{1}{2}\)
Explanation:
E Let the mass of the chain be M. When \(\frac{1}{3}\) of the chain hangs down the edges magnitude of the force on the chain in the downward direction is equal to the weight of \(\frac{1}{3}\) of the chain i.e. \(\mathrm{Mg} / 3\) Then, the mass of chain portion of the chain lying the table is \(2 \mathrm{M} / 3\). The magnitude of normal reaction is \(2 \mathrm{Mg} / 3\). \(\therefore\) Maximum force of friction on the chain \(=\mu_{\mathrm{s}}\) \((2 \mathrm{Mg} / 3)\). When the chain just starts sliding Force on the chain hangs down the edges \(=\) Force on the chain lying the table \(\therefore \quad \mathrm{Mg} / 3=\mu_{\mathrm{s}}(2 \mathrm{Mg} / 3)\) \(\mu_{\mathrm{s}}=\frac{1}{2}\)
