NEET Test Series from KOTA - 10 Papers In MS WORD
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Motion in One Dimensions
141895
When a ball is thrown up vertically with a velocity \(v_{0}\) it reaches a height \(h\). If one wishes to triple the maximum height, then the ball should be thrown with a velocity of
1 \(\sqrt{3} \mathrm{v}_{0}\)
2 \(3 \mathrm{v}_{0}\)
3 \(9 \mathrm{v}_{0}\)
4 \(3 / 2 v_{0}\)
Explanation:
A Here initial velocity, \(u=v_{0}\) Height reached \(=\mathrm{h}\) At highest point final velocity \((\mathrm{v})\) will be zero. From equation of motion, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2\) as \(0-\mathrm{v}_{0}^{2}=-2 \mathrm{gh} \Rightarrow \mathrm{h}=\frac{\mathrm{v}_{0}^{2}}{2 \mathrm{~g}}\) For thrice the maximum height ( \(3 \mathrm{~h})\), \(0-\mathrm{v}_{0}^{\prime 2}=-2 \mathrm{~g}\left(\frac{3 \mathrm{v}_{0}^{2}}{2 \mathrm{~g}}\right)\) \(\quad\left(\quad \left(\mathrm{v}_{0}^{\prime}=\right.\right.\text { new initial velocity) }\) \(\mathrm{v}_{0}^{\prime 2}=3 \mathrm{v}_{0}^{2} \Rightarrow \mathrm{v}_{0}^{\prime}=\sqrt{3} \mathrm{v}_{0}\)
SRMJEE - 2016
Motion in One Dimensions
141897
A stone released with zero velocity from the top of a tower, reaches the ground in \(4 \mathrm{~s}\). The height of the tower is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(20 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(80 \mathrm{~m}\)
4 \(160 \mathrm{~m}\)
Explanation:
C Given that, \(\mathrm{t}=4 \mathrm{sec}, \quad \mathrm{u}=0 \quad \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) (From the equation of motion) \(\mathrm{s}=\mathrm{ut}-\frac{1}{2} \mathrm{gt}^{2}\) \(\mathrm{~s}=0 \times 4-\frac{1}{2} \times 10 \times 4 \times 4\) \(\mathrm{~s}=80 \mathrm{~m}\)
AIPMT 1995
Motion in One Dimensions
141898
If a ball is thrown vertically upwards with a velocity of \(40 \mathrm{~m} / \mathrm{s}\), then velocity of the ball after \(2 \mathrm{~s}\) will be \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(15 \mathrm{~m} / \mathrm{s}\)
2 \(20 \mathrm{~m} / \mathrm{s}\)
3 \(25 \mathrm{~m} / \mathrm{s}\)
4 \(28 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given that, \(\mathrm{u}=40 \mathrm{~m} / \mathrm{s}, \mathrm{t}=2 \mathrm{~s}\) \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) \(\mathrm{a}=-\mathrm{g}\) \(\mathrm{v}=\mathrm{u}+\) at \(\mathrm{v}=40+(-10) \times 2\) \(\mathrm{v}=40-20 \Rightarrow \mathrm{v}=20 \mathrm{~m} / \mathrm{s}\)
AIPMT 1996
Motion in One Dimensions
141899
A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is \(10 \mathrm{~m} / \mathrm{s}\), then the maximum height attained by the stone is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(8 \mathrm{~m}\)
2 \(10 \mathrm{~m}\)
3 \(15 \mathrm{~m}\)
4 \(20 \mathrm{~m}\)
Explanation:
B \(v^{2}-u^{2}=2 g h\) \(0-\mathrm{u}^{2}=2 \mathrm{gh}\) \(\therefore \quad \mathrm{u}=\sqrt{2 \mathrm{gh}}\) At \(\mathrm{h} / 2, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) \(10^{2}-\mathrm{u}^{2}=-2 \mathrm{gh} / 2\) \(100-2 \mathrm{gh}=-\mathrm{gh}\) \(\mathrm{h}=\frac{100}{10}=10 \mathrm{~m}\)
141895
When a ball is thrown up vertically with a velocity \(v_{0}\) it reaches a height \(h\). If one wishes to triple the maximum height, then the ball should be thrown with a velocity of
1 \(\sqrt{3} \mathrm{v}_{0}\)
2 \(3 \mathrm{v}_{0}\)
3 \(9 \mathrm{v}_{0}\)
4 \(3 / 2 v_{0}\)
Explanation:
A Here initial velocity, \(u=v_{0}\) Height reached \(=\mathrm{h}\) At highest point final velocity \((\mathrm{v})\) will be zero. From equation of motion, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2\) as \(0-\mathrm{v}_{0}^{2}=-2 \mathrm{gh} \Rightarrow \mathrm{h}=\frac{\mathrm{v}_{0}^{2}}{2 \mathrm{~g}}\) For thrice the maximum height ( \(3 \mathrm{~h})\), \(0-\mathrm{v}_{0}^{\prime 2}=-2 \mathrm{~g}\left(\frac{3 \mathrm{v}_{0}^{2}}{2 \mathrm{~g}}\right)\) \(\quad\left(\quad \left(\mathrm{v}_{0}^{\prime}=\right.\right.\text { new initial velocity) }\) \(\mathrm{v}_{0}^{\prime 2}=3 \mathrm{v}_{0}^{2} \Rightarrow \mathrm{v}_{0}^{\prime}=\sqrt{3} \mathrm{v}_{0}\)
SRMJEE - 2016
Motion in One Dimensions
141897
A stone released with zero velocity from the top of a tower, reaches the ground in \(4 \mathrm{~s}\). The height of the tower is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(20 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(80 \mathrm{~m}\)
4 \(160 \mathrm{~m}\)
Explanation:
C Given that, \(\mathrm{t}=4 \mathrm{sec}, \quad \mathrm{u}=0 \quad \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) (From the equation of motion) \(\mathrm{s}=\mathrm{ut}-\frac{1}{2} \mathrm{gt}^{2}\) \(\mathrm{~s}=0 \times 4-\frac{1}{2} \times 10 \times 4 \times 4\) \(\mathrm{~s}=80 \mathrm{~m}\)
AIPMT 1995
Motion in One Dimensions
141898
If a ball is thrown vertically upwards with a velocity of \(40 \mathrm{~m} / \mathrm{s}\), then velocity of the ball after \(2 \mathrm{~s}\) will be \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(15 \mathrm{~m} / \mathrm{s}\)
2 \(20 \mathrm{~m} / \mathrm{s}\)
3 \(25 \mathrm{~m} / \mathrm{s}\)
4 \(28 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given that, \(\mathrm{u}=40 \mathrm{~m} / \mathrm{s}, \mathrm{t}=2 \mathrm{~s}\) \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) \(\mathrm{a}=-\mathrm{g}\) \(\mathrm{v}=\mathrm{u}+\) at \(\mathrm{v}=40+(-10) \times 2\) \(\mathrm{v}=40-20 \Rightarrow \mathrm{v}=20 \mathrm{~m} / \mathrm{s}\)
AIPMT 1996
Motion in One Dimensions
141899
A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is \(10 \mathrm{~m} / \mathrm{s}\), then the maximum height attained by the stone is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(8 \mathrm{~m}\)
2 \(10 \mathrm{~m}\)
3 \(15 \mathrm{~m}\)
4 \(20 \mathrm{~m}\)
Explanation:
B \(v^{2}-u^{2}=2 g h\) \(0-\mathrm{u}^{2}=2 \mathrm{gh}\) \(\therefore \quad \mathrm{u}=\sqrt{2 \mathrm{gh}}\) At \(\mathrm{h} / 2, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) \(10^{2}-\mathrm{u}^{2}=-2 \mathrm{gh} / 2\) \(100-2 \mathrm{gh}=-\mathrm{gh}\) \(\mathrm{h}=\frac{100}{10}=10 \mathrm{~m}\)
141895
When a ball is thrown up vertically with a velocity \(v_{0}\) it reaches a height \(h\). If one wishes to triple the maximum height, then the ball should be thrown with a velocity of
1 \(\sqrt{3} \mathrm{v}_{0}\)
2 \(3 \mathrm{v}_{0}\)
3 \(9 \mathrm{v}_{0}\)
4 \(3 / 2 v_{0}\)
Explanation:
A Here initial velocity, \(u=v_{0}\) Height reached \(=\mathrm{h}\) At highest point final velocity \((\mathrm{v})\) will be zero. From equation of motion, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2\) as \(0-\mathrm{v}_{0}^{2}=-2 \mathrm{gh} \Rightarrow \mathrm{h}=\frac{\mathrm{v}_{0}^{2}}{2 \mathrm{~g}}\) For thrice the maximum height ( \(3 \mathrm{~h})\), \(0-\mathrm{v}_{0}^{\prime 2}=-2 \mathrm{~g}\left(\frac{3 \mathrm{v}_{0}^{2}}{2 \mathrm{~g}}\right)\) \(\quad\left(\quad \left(\mathrm{v}_{0}^{\prime}=\right.\right.\text { new initial velocity) }\) \(\mathrm{v}_{0}^{\prime 2}=3 \mathrm{v}_{0}^{2} \Rightarrow \mathrm{v}_{0}^{\prime}=\sqrt{3} \mathrm{v}_{0}\)
SRMJEE - 2016
Motion in One Dimensions
141897
A stone released with zero velocity from the top of a tower, reaches the ground in \(4 \mathrm{~s}\). The height of the tower is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(20 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(80 \mathrm{~m}\)
4 \(160 \mathrm{~m}\)
Explanation:
C Given that, \(\mathrm{t}=4 \mathrm{sec}, \quad \mathrm{u}=0 \quad \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) (From the equation of motion) \(\mathrm{s}=\mathrm{ut}-\frac{1}{2} \mathrm{gt}^{2}\) \(\mathrm{~s}=0 \times 4-\frac{1}{2} \times 10 \times 4 \times 4\) \(\mathrm{~s}=80 \mathrm{~m}\)
AIPMT 1995
Motion in One Dimensions
141898
If a ball is thrown vertically upwards with a velocity of \(40 \mathrm{~m} / \mathrm{s}\), then velocity of the ball after \(2 \mathrm{~s}\) will be \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(15 \mathrm{~m} / \mathrm{s}\)
2 \(20 \mathrm{~m} / \mathrm{s}\)
3 \(25 \mathrm{~m} / \mathrm{s}\)
4 \(28 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given that, \(\mathrm{u}=40 \mathrm{~m} / \mathrm{s}, \mathrm{t}=2 \mathrm{~s}\) \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) \(\mathrm{a}=-\mathrm{g}\) \(\mathrm{v}=\mathrm{u}+\) at \(\mathrm{v}=40+(-10) \times 2\) \(\mathrm{v}=40-20 \Rightarrow \mathrm{v}=20 \mathrm{~m} / \mathrm{s}\)
AIPMT 1996
Motion in One Dimensions
141899
A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is \(10 \mathrm{~m} / \mathrm{s}\), then the maximum height attained by the stone is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(8 \mathrm{~m}\)
2 \(10 \mathrm{~m}\)
3 \(15 \mathrm{~m}\)
4 \(20 \mathrm{~m}\)
Explanation:
B \(v^{2}-u^{2}=2 g h\) \(0-\mathrm{u}^{2}=2 \mathrm{gh}\) \(\therefore \quad \mathrm{u}=\sqrt{2 \mathrm{gh}}\) At \(\mathrm{h} / 2, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) \(10^{2}-\mathrm{u}^{2}=-2 \mathrm{gh} / 2\) \(100-2 \mathrm{gh}=-\mathrm{gh}\) \(\mathrm{h}=\frac{100}{10}=10 \mathrm{~m}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Motion in One Dimensions
141895
When a ball is thrown up vertically with a velocity \(v_{0}\) it reaches a height \(h\). If one wishes to triple the maximum height, then the ball should be thrown with a velocity of
1 \(\sqrt{3} \mathrm{v}_{0}\)
2 \(3 \mathrm{v}_{0}\)
3 \(9 \mathrm{v}_{0}\)
4 \(3 / 2 v_{0}\)
Explanation:
A Here initial velocity, \(u=v_{0}\) Height reached \(=\mathrm{h}\) At highest point final velocity \((\mathrm{v})\) will be zero. From equation of motion, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2\) as \(0-\mathrm{v}_{0}^{2}=-2 \mathrm{gh} \Rightarrow \mathrm{h}=\frac{\mathrm{v}_{0}^{2}}{2 \mathrm{~g}}\) For thrice the maximum height ( \(3 \mathrm{~h})\), \(0-\mathrm{v}_{0}^{\prime 2}=-2 \mathrm{~g}\left(\frac{3 \mathrm{v}_{0}^{2}}{2 \mathrm{~g}}\right)\) \(\quad\left(\quad \left(\mathrm{v}_{0}^{\prime}=\right.\right.\text { new initial velocity) }\) \(\mathrm{v}_{0}^{\prime 2}=3 \mathrm{v}_{0}^{2} \Rightarrow \mathrm{v}_{0}^{\prime}=\sqrt{3} \mathrm{v}_{0}\)
SRMJEE - 2016
Motion in One Dimensions
141897
A stone released with zero velocity from the top of a tower, reaches the ground in \(4 \mathrm{~s}\). The height of the tower is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(20 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(80 \mathrm{~m}\)
4 \(160 \mathrm{~m}\)
Explanation:
C Given that, \(\mathrm{t}=4 \mathrm{sec}, \quad \mathrm{u}=0 \quad \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) (From the equation of motion) \(\mathrm{s}=\mathrm{ut}-\frac{1}{2} \mathrm{gt}^{2}\) \(\mathrm{~s}=0 \times 4-\frac{1}{2} \times 10 \times 4 \times 4\) \(\mathrm{~s}=80 \mathrm{~m}\)
AIPMT 1995
Motion in One Dimensions
141898
If a ball is thrown vertically upwards with a velocity of \(40 \mathrm{~m} / \mathrm{s}\), then velocity of the ball after \(2 \mathrm{~s}\) will be \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(15 \mathrm{~m} / \mathrm{s}\)
2 \(20 \mathrm{~m} / \mathrm{s}\)
3 \(25 \mathrm{~m} / \mathrm{s}\)
4 \(28 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given that, \(\mathrm{u}=40 \mathrm{~m} / \mathrm{s}, \mathrm{t}=2 \mathrm{~s}\) \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) \(\mathrm{a}=-\mathrm{g}\) \(\mathrm{v}=\mathrm{u}+\) at \(\mathrm{v}=40+(-10) \times 2\) \(\mathrm{v}=40-20 \Rightarrow \mathrm{v}=20 \mathrm{~m} / \mathrm{s}\)
AIPMT 1996
Motion in One Dimensions
141899
A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is \(10 \mathrm{~m} / \mathrm{s}\), then the maximum height attained by the stone is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(8 \mathrm{~m}\)
2 \(10 \mathrm{~m}\)
3 \(15 \mathrm{~m}\)
4 \(20 \mathrm{~m}\)
Explanation:
B \(v^{2}-u^{2}=2 g h\) \(0-\mathrm{u}^{2}=2 \mathrm{gh}\) \(\therefore \quad \mathrm{u}=\sqrt{2 \mathrm{gh}}\) At \(\mathrm{h} / 2, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) \(10^{2}-\mathrm{u}^{2}=-2 \mathrm{gh} / 2\) \(100-2 \mathrm{gh}=-\mathrm{gh}\) \(\mathrm{h}=\frac{100}{10}=10 \mathrm{~m}\)