141970
A smooth inclined plane is inclined at an angle \(\theta\) with horizontal. A body starts from rest and slides down the inclined surface. Then the time taken by it to reach the bottom is
C Given, the initial velocity of body will be zero because body moves from rest i.e. \(\mathrm{u}=0\) According to second law of motion- \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Here, \(\quad \mathrm{s}=l, \quad \mathrm{a}=\mathrm{g} \sin \theta\) \(\therefore \quad l=0 \times \mathrm{t}+\frac{1}{2}(\mathrm{~g} \sin \theta) \times \mathrm{t}^{2}\) \(\mathrm{t}=\sqrt{\frac{2 l}{\mathrm{~g} \sin \theta}}\) In \(\triangle \mathrm{BAC}\) - \(\sin \theta =\frac{\mathrm{h}}{l}\) \(l =\frac{\mathrm{h}}{\sin \theta}\) Putting the value of \(l\) in equation (i), we get- \(\mathrm{t}=\sqrt{\frac{\frac{2 \times h}{\sin \theta}}{\mathrm{g} \sin \theta}}\) or \(\quad \mathrm{t}=\frac{1}{\sin \theta} \sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\)
AIIMS-2015
Motion in One Dimensions
141971
A block of mass \(m\) is placed on a smooth wedge of inclination \(\theta\). The whole system is accelerated horizontally, so that the block does not slip on the wedge. The force exerted by the wedge on the block ( \(\mathrm{g}\) is acceleration due to gravity) will be
1 \(\mathrm{mg} \cos \theta\)
2 \(m g \sin \theta\)
3 \(\mathrm{mg}\)
4 \(\frac{\mathrm{mg}}{\cos \theta}\)
Explanation:
D Given, mass \(=\mathrm{m}\), angle \(=\theta\) According to diagram, \(\mathrm{mg} \sin \theta=\mathrm{ma} \cos \theta \quad\) (along the inclined plane) \(\mathrm{a}=\frac{\mathrm{g} \sin \theta}{\cos \theta}\) \(\mathrm{a}=\mathrm{g} \tan \theta\) Hence, Total reaction of the wedge on the block is \(\mathrm{N}=\mathrm{mg} \cos \theta+\mathrm{ma} \sin \theta\) \(\mathrm{N}=\mathrm{mg} \cos \theta+\frac{\mathrm{mg} \sin \theta \cdot \sin \theta}{\cos \theta}\) \(\mathrm{N}=\frac{\mathrm{mg}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}{\cos \theta} \quad\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right]\) \(\mathrm{N}=\frac{\mathrm{mg}}{\cos \theta}\)
141970
A smooth inclined plane is inclined at an angle \(\theta\) with horizontal. A body starts from rest and slides down the inclined surface. Then the time taken by it to reach the bottom is
C Given, the initial velocity of body will be zero because body moves from rest i.e. \(\mathrm{u}=0\) According to second law of motion- \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Here, \(\quad \mathrm{s}=l, \quad \mathrm{a}=\mathrm{g} \sin \theta\) \(\therefore \quad l=0 \times \mathrm{t}+\frac{1}{2}(\mathrm{~g} \sin \theta) \times \mathrm{t}^{2}\) \(\mathrm{t}=\sqrt{\frac{2 l}{\mathrm{~g} \sin \theta}}\) In \(\triangle \mathrm{BAC}\) - \(\sin \theta =\frac{\mathrm{h}}{l}\) \(l =\frac{\mathrm{h}}{\sin \theta}\) Putting the value of \(l\) in equation (i), we get- \(\mathrm{t}=\sqrt{\frac{\frac{2 \times h}{\sin \theta}}{\mathrm{g} \sin \theta}}\) or \(\quad \mathrm{t}=\frac{1}{\sin \theta} \sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\)
AIIMS-2015
Motion in One Dimensions
141971
A block of mass \(m\) is placed on a smooth wedge of inclination \(\theta\). The whole system is accelerated horizontally, so that the block does not slip on the wedge. The force exerted by the wedge on the block ( \(\mathrm{g}\) is acceleration due to gravity) will be
1 \(\mathrm{mg} \cos \theta\)
2 \(m g \sin \theta\)
3 \(\mathrm{mg}\)
4 \(\frac{\mathrm{mg}}{\cos \theta}\)
Explanation:
D Given, mass \(=\mathrm{m}\), angle \(=\theta\) According to diagram, \(\mathrm{mg} \sin \theta=\mathrm{ma} \cos \theta \quad\) (along the inclined plane) \(\mathrm{a}=\frac{\mathrm{g} \sin \theta}{\cos \theta}\) \(\mathrm{a}=\mathrm{g} \tan \theta\) Hence, Total reaction of the wedge on the block is \(\mathrm{N}=\mathrm{mg} \cos \theta+\mathrm{ma} \sin \theta\) \(\mathrm{N}=\mathrm{mg} \cos \theta+\frac{\mathrm{mg} \sin \theta \cdot \sin \theta}{\cos \theta}\) \(\mathrm{N}=\frac{\mathrm{mg}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}{\cos \theta} \quad\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right]\) \(\mathrm{N}=\frac{\mathrm{mg}}{\cos \theta}\)
