362401
A body standing at the top of a tower of \(20 \, m\) height drops a stone. Assuming, \(g = \,10\,m{s^{ - 2}},\) the velocity with which it hits the ground is
1 \(40\,m/s\)
2 \(20\,m/s\)
3 \(5\,m/s\)
4 \(10\,m/s\)
Explanation:
Given \(g = \,10\,m{s^{ - 2}}\,{\rm{and}}\,h = 20 \, m\) We have \(v = \sqrt {2gh} \) \( = \sqrt {2 \times 10 \times 20} = \sqrt {400} = 20\,m/s\)
PHXI03:MOTION IN A STRAIGHT LINE
362402
Assertion : If air resistance is considered, then time of ascent and time of descent will be different. Reason : Magnitudes of acceleration will be different in upward and downward motion.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
When air resistance is considered, it affects the motion of the object differently during ascent and descent, leading to different times of ascent and descent. When air resistance is considered, the magnitude of acceleration due to gravity is affected by the opposing force of air resistance. This can result in different accelerations during upward and downward motion. The reason provides a correct explanation for why the assertion is true. So correct option is (1)
PHXI03:MOTION IN A STRAIGHT LINE
362403
\({B_1},{B_2}\,{\rm{and}}\,{B_3}\) are three balloons ascending with velocities \(v,2v\,{\rm{and}}\,3v\), respectively. If a bomb is dropped from each balloon when they are at the same height, then
1 Bomb from \({B_1}\) reaches ground first
2 Bomb from \({B_2}\) reaches ground first
3 Bomb from \({B_3}\) reaches ground first
4 They reach the ground simultaneously
Explanation:
Let \(h\) is the initial height of the balloons and \(u\) be the initial velocity of the bomb with respect to ground. \( - h = ut - \frac{1}{2}g{t^2} \Rightarrow h = \frac{1}{2}g{t^2} - ut\) \( \Rightarrow \frac{1}{2}g{t^2} - ut - h = 0\) \(t = \frac{{u + \sqrt {{u^2} + 2gh} }}{g}\) So as \(u\) increases, \(t\) increases.
PHXI03:MOTION IN A STRAIGHT LINE
362404
A ball is released from the top of a tower of height \(h\) meters. It takes \(T\) seconds to reach the ground. What is the position of the ball at \(\frac{T}{3}\)second
We have, \(s = ut + \frac{1}{2}g{t^2}\) \( \Rightarrow {\rm{ }}h = \frac{1}{2}g{T^2}\,\,\left( {\therefore u = 0} \right)\) In \(T/3\) second, vertical distance moved is given by \(h' = \frac{1}{2}g{\left( {\frac{T}{3}} \right)^2} \Rightarrow h' = \frac{1}{2} \times \frac{{g{T^2}}}{9} = \frac{h}{9}\) \(\therefore \) Position of ball from ground \( = h - \frac{h}{9} = \frac{{8h}}{9}\)
362401
A body standing at the top of a tower of \(20 \, m\) height drops a stone. Assuming, \(g = \,10\,m{s^{ - 2}},\) the velocity with which it hits the ground is
1 \(40\,m/s\)
2 \(20\,m/s\)
3 \(5\,m/s\)
4 \(10\,m/s\)
Explanation:
Given \(g = \,10\,m{s^{ - 2}}\,{\rm{and}}\,h = 20 \, m\) We have \(v = \sqrt {2gh} \) \( = \sqrt {2 \times 10 \times 20} = \sqrt {400} = 20\,m/s\)
PHXI03:MOTION IN A STRAIGHT LINE
362402
Assertion : If air resistance is considered, then time of ascent and time of descent will be different. Reason : Magnitudes of acceleration will be different in upward and downward motion.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
When air resistance is considered, it affects the motion of the object differently during ascent and descent, leading to different times of ascent and descent. When air resistance is considered, the magnitude of acceleration due to gravity is affected by the opposing force of air resistance. This can result in different accelerations during upward and downward motion. The reason provides a correct explanation for why the assertion is true. So correct option is (1)
PHXI03:MOTION IN A STRAIGHT LINE
362403
\({B_1},{B_2}\,{\rm{and}}\,{B_3}\) are three balloons ascending with velocities \(v,2v\,{\rm{and}}\,3v\), respectively. If a bomb is dropped from each balloon when they are at the same height, then
1 Bomb from \({B_1}\) reaches ground first
2 Bomb from \({B_2}\) reaches ground first
3 Bomb from \({B_3}\) reaches ground first
4 They reach the ground simultaneously
Explanation:
Let \(h\) is the initial height of the balloons and \(u\) be the initial velocity of the bomb with respect to ground. \( - h = ut - \frac{1}{2}g{t^2} \Rightarrow h = \frac{1}{2}g{t^2} - ut\) \( \Rightarrow \frac{1}{2}g{t^2} - ut - h = 0\) \(t = \frac{{u + \sqrt {{u^2} + 2gh} }}{g}\) So as \(u\) increases, \(t\) increases.
PHXI03:MOTION IN A STRAIGHT LINE
362404
A ball is released from the top of a tower of height \(h\) meters. It takes \(T\) seconds to reach the ground. What is the position of the ball at \(\frac{T}{3}\)second
We have, \(s = ut + \frac{1}{2}g{t^2}\) \( \Rightarrow {\rm{ }}h = \frac{1}{2}g{T^2}\,\,\left( {\therefore u = 0} \right)\) In \(T/3\) second, vertical distance moved is given by \(h' = \frac{1}{2}g{\left( {\frac{T}{3}} \right)^2} \Rightarrow h' = \frac{1}{2} \times \frac{{g{T^2}}}{9} = \frac{h}{9}\) \(\therefore \) Position of ball from ground \( = h - \frac{h}{9} = \frac{{8h}}{9}\)
362401
A body standing at the top of a tower of \(20 \, m\) height drops a stone. Assuming, \(g = \,10\,m{s^{ - 2}},\) the velocity with which it hits the ground is
1 \(40\,m/s\)
2 \(20\,m/s\)
3 \(5\,m/s\)
4 \(10\,m/s\)
Explanation:
Given \(g = \,10\,m{s^{ - 2}}\,{\rm{and}}\,h = 20 \, m\) We have \(v = \sqrt {2gh} \) \( = \sqrt {2 \times 10 \times 20} = \sqrt {400} = 20\,m/s\)
PHXI03:MOTION IN A STRAIGHT LINE
362402
Assertion : If air resistance is considered, then time of ascent and time of descent will be different. Reason : Magnitudes of acceleration will be different in upward and downward motion.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
When air resistance is considered, it affects the motion of the object differently during ascent and descent, leading to different times of ascent and descent. When air resistance is considered, the magnitude of acceleration due to gravity is affected by the opposing force of air resistance. This can result in different accelerations during upward and downward motion. The reason provides a correct explanation for why the assertion is true. So correct option is (1)
PHXI03:MOTION IN A STRAIGHT LINE
362403
\({B_1},{B_2}\,{\rm{and}}\,{B_3}\) are three balloons ascending with velocities \(v,2v\,{\rm{and}}\,3v\), respectively. If a bomb is dropped from each balloon when they are at the same height, then
1 Bomb from \({B_1}\) reaches ground first
2 Bomb from \({B_2}\) reaches ground first
3 Bomb from \({B_3}\) reaches ground first
4 They reach the ground simultaneously
Explanation:
Let \(h\) is the initial height of the balloons and \(u\) be the initial velocity of the bomb with respect to ground. \( - h = ut - \frac{1}{2}g{t^2} \Rightarrow h = \frac{1}{2}g{t^2} - ut\) \( \Rightarrow \frac{1}{2}g{t^2} - ut - h = 0\) \(t = \frac{{u + \sqrt {{u^2} + 2gh} }}{g}\) So as \(u\) increases, \(t\) increases.
PHXI03:MOTION IN A STRAIGHT LINE
362404
A ball is released from the top of a tower of height \(h\) meters. It takes \(T\) seconds to reach the ground. What is the position of the ball at \(\frac{T}{3}\)second
We have, \(s = ut + \frac{1}{2}g{t^2}\) \( \Rightarrow {\rm{ }}h = \frac{1}{2}g{T^2}\,\,\left( {\therefore u = 0} \right)\) In \(T/3\) second, vertical distance moved is given by \(h' = \frac{1}{2}g{\left( {\frac{T}{3}} \right)^2} \Rightarrow h' = \frac{1}{2} \times \frac{{g{T^2}}}{9} = \frac{h}{9}\) \(\therefore \) Position of ball from ground \( = h - \frac{h}{9} = \frac{{8h}}{9}\)
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PHXI03:MOTION IN A STRAIGHT LINE
362401
A body standing at the top of a tower of \(20 \, m\) height drops a stone. Assuming, \(g = \,10\,m{s^{ - 2}},\) the velocity with which it hits the ground is
1 \(40\,m/s\)
2 \(20\,m/s\)
3 \(5\,m/s\)
4 \(10\,m/s\)
Explanation:
Given \(g = \,10\,m{s^{ - 2}}\,{\rm{and}}\,h = 20 \, m\) We have \(v = \sqrt {2gh} \) \( = \sqrt {2 \times 10 \times 20} = \sqrt {400} = 20\,m/s\)
PHXI03:MOTION IN A STRAIGHT LINE
362402
Assertion : If air resistance is considered, then time of ascent and time of descent will be different. Reason : Magnitudes of acceleration will be different in upward and downward motion.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
When air resistance is considered, it affects the motion of the object differently during ascent and descent, leading to different times of ascent and descent. When air resistance is considered, the magnitude of acceleration due to gravity is affected by the opposing force of air resistance. This can result in different accelerations during upward and downward motion. The reason provides a correct explanation for why the assertion is true. So correct option is (1)
PHXI03:MOTION IN A STRAIGHT LINE
362403
\({B_1},{B_2}\,{\rm{and}}\,{B_3}\) are three balloons ascending with velocities \(v,2v\,{\rm{and}}\,3v\), respectively. If a bomb is dropped from each balloon when they are at the same height, then
1 Bomb from \({B_1}\) reaches ground first
2 Bomb from \({B_2}\) reaches ground first
3 Bomb from \({B_3}\) reaches ground first
4 They reach the ground simultaneously
Explanation:
Let \(h\) is the initial height of the balloons and \(u\) be the initial velocity of the bomb with respect to ground. \( - h = ut - \frac{1}{2}g{t^2} \Rightarrow h = \frac{1}{2}g{t^2} - ut\) \( \Rightarrow \frac{1}{2}g{t^2} - ut - h = 0\) \(t = \frac{{u + \sqrt {{u^2} + 2gh} }}{g}\) So as \(u\) increases, \(t\) increases.
PHXI03:MOTION IN A STRAIGHT LINE
362404
A ball is released from the top of a tower of height \(h\) meters. It takes \(T\) seconds to reach the ground. What is the position of the ball at \(\frac{T}{3}\)second
We have, \(s = ut + \frac{1}{2}g{t^2}\) \( \Rightarrow {\rm{ }}h = \frac{1}{2}g{T^2}\,\,\left( {\therefore u = 0} \right)\) In \(T/3\) second, vertical distance moved is given by \(h' = \frac{1}{2}g{\left( {\frac{T}{3}} \right)^2} \Rightarrow h' = \frac{1}{2} \times \frac{{g{T^2}}}{9} = \frac{h}{9}\) \(\therefore \) Position of ball from ground \( = h - \frac{h}{9} = \frac{{8h}}{9}\)
