NEET Test Series from KOTA - 10 Papers In MS WORD
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LAWS OF MOTION (ADDITIONAL)
372107
An object placed on an inclined plane starts sliding when the angle of incline becomes \(30^{\circ}\). The coefficient of static friction between the object and the plane is
1 \(\frac{1}{\sqrt{3}}\)
2 \(\sqrt{3}\)
3 \(\frac{1}{2}\)
4 \(\frac{\sqrt{3}}{2}\)
Explanation:
A Inclination angle, \(\theta=30^{\circ}\) Coefficient of static friction \(\mu_{\mathrm{s}}=\) ? We know that, \(\mu_{\mathrm{s}}=\tan \theta\) \(\therefore \quad \mu_{\mathrm{s}}=\tan 30^{\circ}\) \(\mu_{\mathrm{s}}=\frac{1}{\sqrt{3}}\) Hence, coefficient of static friction, \(\mu_{\mathrm{s}}=\frac{1}{\sqrt{3}}\)
J and K CET- 2010
LAWS OF MOTION (ADDITIONAL)
372108
With increase of temperature, the frictional force acting between two surfaces
1 increases
2 remains the same
3 decreases
4 becomes zero
Explanation:
B Friction, which acts between the contact of two surfaces, it means friction is the function of roughness of surface and force acting on it. With increase in temperature the frictional force acting between the two surface remains the same.
J and K CET- 2007
LAWS OF MOTION (ADDITIONAL)
372109
A block of mass \(2 \mathrm{~kg}\) rests on a horizontal surface. If a horizontal force of \(5 \mathrm{~N}\) is applied on the block the frictional force on it is \(\left(\mu_{\mathrm{k}}=0.4, \mu_{\mathrm{s}}=0.5\right)\)
1 \(5 \mathrm{~N}\)
2 \(10 \mathrm{~N}\)
3 8
4 zero
Explanation:
A Given, block mass (m) \(=2 \mathrm{~kg}\) Horizontal force \((\mathrm{F})=5 \mathrm{~N}\) Frictional force \(\left(\mathrm{f}_{\mathrm{s}}\right)=\) ? From FBD \(\mathrm{N}=\mathrm{mg}\) \(\mathrm{f}=\mathrm{F}_{\mathrm{net}}\) The frictional force is calculated using the formula - \(\mathrm{f}_{\mathrm{s}}=\mu_{\mathrm{s}} \mathrm{N}\) \(\mathrm{f}_{\mathrm{s}}=\mu_{\mathrm{s}} \mathrm{mg}=0.5 \times 2 \times 10\) \(\mathrm{f}_{\mathrm{s}}=10 \mathrm{~N}\) Here, we see that the frictional force is greater than the applied force, i.e. \(\mathrm{f}_{\mathrm{s}}>\mathrm{F}\) So, frictional force will adjust its value and will be equal to applied horizontal force Frictional force, \(\mathrm{f}=5 \mathrm{~N}\)
J and K CET- 2006
LAWS OF MOTION (ADDITIONAL)
372110
An inclined plane has an inclination \(\theta\) with horizontal. A body of mass \(m\) rests on it. If the coefficient of friction between the body and the plane is \(\mu\). Then the minimum force that needs to be applied parallel to the inclined plane is
C Given, Inclination \(=\theta\) with horizontal mass \(=\mathrm{m}\) at rest Coefficient of friction \(=\mu\) Minimum force, parallel to the inclined plane \(=\) ? From FBD Net force \(\mathrm{f}_{\mathrm{f}}=\mu \mathrm{N}=\mu \mathrm{mg} \cos \theta\) \(\mathrm{f}_{\text {net }}=\mathrm{f}_{\mathrm{f}}+\mathrm{f}\) \(\mathrm{f}_{\text {net }}=\mu \mathrm{mg} \cos \theta+\mathrm{mg} \sin \theta\)
372107
An object placed on an inclined plane starts sliding when the angle of incline becomes \(30^{\circ}\). The coefficient of static friction between the object and the plane is
1 \(\frac{1}{\sqrt{3}}\)
2 \(\sqrt{3}\)
3 \(\frac{1}{2}\)
4 \(\frac{\sqrt{3}}{2}\)
Explanation:
A Inclination angle, \(\theta=30^{\circ}\) Coefficient of static friction \(\mu_{\mathrm{s}}=\) ? We know that, \(\mu_{\mathrm{s}}=\tan \theta\) \(\therefore \quad \mu_{\mathrm{s}}=\tan 30^{\circ}\) \(\mu_{\mathrm{s}}=\frac{1}{\sqrt{3}}\) Hence, coefficient of static friction, \(\mu_{\mathrm{s}}=\frac{1}{\sqrt{3}}\)
J and K CET- 2010
LAWS OF MOTION (ADDITIONAL)
372108
With increase of temperature, the frictional force acting between two surfaces
1 increases
2 remains the same
3 decreases
4 becomes zero
Explanation:
B Friction, which acts between the contact of two surfaces, it means friction is the function of roughness of surface and force acting on it. With increase in temperature the frictional force acting between the two surface remains the same.
J and K CET- 2007
LAWS OF MOTION (ADDITIONAL)
372109
A block of mass \(2 \mathrm{~kg}\) rests on a horizontal surface. If a horizontal force of \(5 \mathrm{~N}\) is applied on the block the frictional force on it is \(\left(\mu_{\mathrm{k}}=0.4, \mu_{\mathrm{s}}=0.5\right)\)
1 \(5 \mathrm{~N}\)
2 \(10 \mathrm{~N}\)
3 8
4 zero
Explanation:
A Given, block mass (m) \(=2 \mathrm{~kg}\) Horizontal force \((\mathrm{F})=5 \mathrm{~N}\) Frictional force \(\left(\mathrm{f}_{\mathrm{s}}\right)=\) ? From FBD \(\mathrm{N}=\mathrm{mg}\) \(\mathrm{f}=\mathrm{F}_{\mathrm{net}}\) The frictional force is calculated using the formula - \(\mathrm{f}_{\mathrm{s}}=\mu_{\mathrm{s}} \mathrm{N}\) \(\mathrm{f}_{\mathrm{s}}=\mu_{\mathrm{s}} \mathrm{mg}=0.5 \times 2 \times 10\) \(\mathrm{f}_{\mathrm{s}}=10 \mathrm{~N}\) Here, we see that the frictional force is greater than the applied force, i.e. \(\mathrm{f}_{\mathrm{s}}>\mathrm{F}\) So, frictional force will adjust its value and will be equal to applied horizontal force Frictional force, \(\mathrm{f}=5 \mathrm{~N}\)
J and K CET- 2006
LAWS OF MOTION (ADDITIONAL)
372110
An inclined plane has an inclination \(\theta\) with horizontal. A body of mass \(m\) rests on it. If the coefficient of friction between the body and the plane is \(\mu\). Then the minimum force that needs to be applied parallel to the inclined plane is
C Given, Inclination \(=\theta\) with horizontal mass \(=\mathrm{m}\) at rest Coefficient of friction \(=\mu\) Minimum force, parallel to the inclined plane \(=\) ? From FBD Net force \(\mathrm{f}_{\mathrm{f}}=\mu \mathrm{N}=\mu \mathrm{mg} \cos \theta\) \(\mathrm{f}_{\text {net }}=\mathrm{f}_{\mathrm{f}}+\mathrm{f}\) \(\mathrm{f}_{\text {net }}=\mu \mathrm{mg} \cos \theta+\mathrm{mg} \sin \theta\)
372107
An object placed on an inclined plane starts sliding when the angle of incline becomes \(30^{\circ}\). The coefficient of static friction between the object and the plane is
1 \(\frac{1}{\sqrt{3}}\)
2 \(\sqrt{3}\)
3 \(\frac{1}{2}\)
4 \(\frac{\sqrt{3}}{2}\)
Explanation:
A Inclination angle, \(\theta=30^{\circ}\) Coefficient of static friction \(\mu_{\mathrm{s}}=\) ? We know that, \(\mu_{\mathrm{s}}=\tan \theta\) \(\therefore \quad \mu_{\mathrm{s}}=\tan 30^{\circ}\) \(\mu_{\mathrm{s}}=\frac{1}{\sqrt{3}}\) Hence, coefficient of static friction, \(\mu_{\mathrm{s}}=\frac{1}{\sqrt{3}}\)
J and K CET- 2010
LAWS OF MOTION (ADDITIONAL)
372108
With increase of temperature, the frictional force acting between two surfaces
1 increases
2 remains the same
3 decreases
4 becomes zero
Explanation:
B Friction, which acts between the contact of two surfaces, it means friction is the function of roughness of surface and force acting on it. With increase in temperature the frictional force acting between the two surface remains the same.
J and K CET- 2007
LAWS OF MOTION (ADDITIONAL)
372109
A block of mass \(2 \mathrm{~kg}\) rests on a horizontal surface. If a horizontal force of \(5 \mathrm{~N}\) is applied on the block the frictional force on it is \(\left(\mu_{\mathrm{k}}=0.4, \mu_{\mathrm{s}}=0.5\right)\)
1 \(5 \mathrm{~N}\)
2 \(10 \mathrm{~N}\)
3 8
4 zero
Explanation:
A Given, block mass (m) \(=2 \mathrm{~kg}\) Horizontal force \((\mathrm{F})=5 \mathrm{~N}\) Frictional force \(\left(\mathrm{f}_{\mathrm{s}}\right)=\) ? From FBD \(\mathrm{N}=\mathrm{mg}\) \(\mathrm{f}=\mathrm{F}_{\mathrm{net}}\) The frictional force is calculated using the formula - \(\mathrm{f}_{\mathrm{s}}=\mu_{\mathrm{s}} \mathrm{N}\) \(\mathrm{f}_{\mathrm{s}}=\mu_{\mathrm{s}} \mathrm{mg}=0.5 \times 2 \times 10\) \(\mathrm{f}_{\mathrm{s}}=10 \mathrm{~N}\) Here, we see that the frictional force is greater than the applied force, i.e. \(\mathrm{f}_{\mathrm{s}}>\mathrm{F}\) So, frictional force will adjust its value and will be equal to applied horizontal force Frictional force, \(\mathrm{f}=5 \mathrm{~N}\)
J and K CET- 2006
LAWS OF MOTION (ADDITIONAL)
372110
An inclined plane has an inclination \(\theta\) with horizontal. A body of mass \(m\) rests on it. If the coefficient of friction between the body and the plane is \(\mu\). Then the minimum force that needs to be applied parallel to the inclined plane is
C Given, Inclination \(=\theta\) with horizontal mass \(=\mathrm{m}\) at rest Coefficient of friction \(=\mu\) Minimum force, parallel to the inclined plane \(=\) ? From FBD Net force \(\mathrm{f}_{\mathrm{f}}=\mu \mathrm{N}=\mu \mathrm{mg} \cos \theta\) \(\mathrm{f}_{\text {net }}=\mathrm{f}_{\mathrm{f}}+\mathrm{f}\) \(\mathrm{f}_{\text {net }}=\mu \mathrm{mg} \cos \theta+\mathrm{mg} \sin \theta\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
LAWS OF MOTION (ADDITIONAL)
372107
An object placed on an inclined plane starts sliding when the angle of incline becomes \(30^{\circ}\). The coefficient of static friction between the object and the plane is
1 \(\frac{1}{\sqrt{3}}\)
2 \(\sqrt{3}\)
3 \(\frac{1}{2}\)
4 \(\frac{\sqrt{3}}{2}\)
Explanation:
A Inclination angle, \(\theta=30^{\circ}\) Coefficient of static friction \(\mu_{\mathrm{s}}=\) ? We know that, \(\mu_{\mathrm{s}}=\tan \theta\) \(\therefore \quad \mu_{\mathrm{s}}=\tan 30^{\circ}\) \(\mu_{\mathrm{s}}=\frac{1}{\sqrt{3}}\) Hence, coefficient of static friction, \(\mu_{\mathrm{s}}=\frac{1}{\sqrt{3}}\)
J and K CET- 2010
LAWS OF MOTION (ADDITIONAL)
372108
With increase of temperature, the frictional force acting between two surfaces
1 increases
2 remains the same
3 decreases
4 becomes zero
Explanation:
B Friction, which acts between the contact of two surfaces, it means friction is the function of roughness of surface and force acting on it. With increase in temperature the frictional force acting between the two surface remains the same.
J and K CET- 2007
LAWS OF MOTION (ADDITIONAL)
372109
A block of mass \(2 \mathrm{~kg}\) rests on a horizontal surface. If a horizontal force of \(5 \mathrm{~N}\) is applied on the block the frictional force on it is \(\left(\mu_{\mathrm{k}}=0.4, \mu_{\mathrm{s}}=0.5\right)\)
1 \(5 \mathrm{~N}\)
2 \(10 \mathrm{~N}\)
3 8
4 zero
Explanation:
A Given, block mass (m) \(=2 \mathrm{~kg}\) Horizontal force \((\mathrm{F})=5 \mathrm{~N}\) Frictional force \(\left(\mathrm{f}_{\mathrm{s}}\right)=\) ? From FBD \(\mathrm{N}=\mathrm{mg}\) \(\mathrm{f}=\mathrm{F}_{\mathrm{net}}\) The frictional force is calculated using the formula - \(\mathrm{f}_{\mathrm{s}}=\mu_{\mathrm{s}} \mathrm{N}\) \(\mathrm{f}_{\mathrm{s}}=\mu_{\mathrm{s}} \mathrm{mg}=0.5 \times 2 \times 10\) \(\mathrm{f}_{\mathrm{s}}=10 \mathrm{~N}\) Here, we see that the frictional force is greater than the applied force, i.e. \(\mathrm{f}_{\mathrm{s}}>\mathrm{F}\) So, frictional force will adjust its value and will be equal to applied horizontal force Frictional force, \(\mathrm{f}=5 \mathrm{~N}\)
J and K CET- 2006
LAWS OF MOTION (ADDITIONAL)
372110
An inclined plane has an inclination \(\theta\) with horizontal. A body of mass \(m\) rests on it. If the coefficient of friction between the body and the plane is \(\mu\). Then the minimum force that needs to be applied parallel to the inclined plane is
C Given, Inclination \(=\theta\) with horizontal mass \(=\mathrm{m}\) at rest Coefficient of friction \(=\mu\) Minimum force, parallel to the inclined plane \(=\) ? From FBD Net force \(\mathrm{f}_{\mathrm{f}}=\mu \mathrm{N}=\mu \mathrm{mg} \cos \theta\) \(\mathrm{f}_{\text {net }}=\mathrm{f}_{\mathrm{f}}+\mathrm{f}\) \(\mathrm{f}_{\text {net }}=\mu \mathrm{mg} \cos \theta+\mathrm{mg} \sin \theta\)
