160800
A ball of mass $25 \mathrm{~g}$, moving with a velocity of $2 \mathrm{~ms}^{-1}$ is stopped within $5 \mathrm{~cm}$, The average resistance offered to the ball is
160801
A batsman deflects a ball by an angle of $90^{\circ}$ without changing its initial speed which is equal to $54 \sqrt{2} \mathbf{~ k m h}^{-1}$. Mass of the ball is $0.15 \mathrm{~kg}$. What is the impulse imparted to the ball?
1 $4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction opposite the initial velocity
2 $4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction of the initial velocity
3 $4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction along the bisector of the initial and final directions of the velocity
4 $45 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction of the final velocity
Explanation:
$4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction along the bisector of the initial and final directions of the velocity
NCERT-I-96
2 RBTS PAPER
160802
A man of mass $70 \mathrm{~kg}$ stands on a weighing scale in a lift which is moving upwards with a uniform speed of $10 \mathrm{~ms}^{-1}$, what would be the reading on the scale?
1 $750 \mathrm{~N}$
2 $700 \mathrm{~N}$
3 $350 \mathrm{~N}$
4 $1050 \mathrm{~N}$
Explanation:
Given that mass of the man, $\mathrm{m}$ is $=70 \mathrm{~kg}$ Acceleration, $\mathrm{a}=0$ Using Newton's second law of motion, we can write the equation of motion as: $\mathrm{R}-\mathrm{mg}=\mathrm{ma}$ Where ma is the net force acting on the man. As the lift is moving at a uniform speed, so the acceleration $\mathrm{a}=0$ $\therefore \quad R=m g=70 \times 10=700 N$
NLI Expert
2 RBTS PAPER
160803
Sand is being dropped from a conveyer belt at the rate of $\mathrm{M} \mathrm{kg} / \mathrm{s}$. The force necessary to keep the belt moving with a constant velocity of $\mathrm{v} \mathrm{m} / \mathrm{s}$ will be
1 2 Mv newton
2 Mv newton
3 $\frac{M v}{2}$ newton
4 zero
Explanation:
$\begin{aligned} F & =\frac{d p}{d t}=\frac{d}{d t}(m v)=v \frac{d m}{d t} \\ & =v \times M=M v \text { newton } \end{aligned}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
2 RBTS PAPER
160800
A ball of mass $25 \mathrm{~g}$, moving with a velocity of $2 \mathrm{~ms}^{-1}$ is stopped within $5 \mathrm{~cm}$, The average resistance offered to the ball is
160801
A batsman deflects a ball by an angle of $90^{\circ}$ without changing its initial speed which is equal to $54 \sqrt{2} \mathbf{~ k m h}^{-1}$. Mass of the ball is $0.15 \mathrm{~kg}$. What is the impulse imparted to the ball?
1 $4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction opposite the initial velocity
2 $4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction of the initial velocity
3 $4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction along the bisector of the initial and final directions of the velocity
4 $45 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction of the final velocity
Explanation:
$4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction along the bisector of the initial and final directions of the velocity
NCERT-I-96
2 RBTS PAPER
160802
A man of mass $70 \mathrm{~kg}$ stands on a weighing scale in a lift which is moving upwards with a uniform speed of $10 \mathrm{~ms}^{-1}$, what would be the reading on the scale?
1 $750 \mathrm{~N}$
2 $700 \mathrm{~N}$
3 $350 \mathrm{~N}$
4 $1050 \mathrm{~N}$
Explanation:
Given that mass of the man, $\mathrm{m}$ is $=70 \mathrm{~kg}$ Acceleration, $\mathrm{a}=0$ Using Newton's second law of motion, we can write the equation of motion as: $\mathrm{R}-\mathrm{mg}=\mathrm{ma}$ Where ma is the net force acting on the man. As the lift is moving at a uniform speed, so the acceleration $\mathrm{a}=0$ $\therefore \quad R=m g=70 \times 10=700 N$
NLI Expert
2 RBTS PAPER
160803
Sand is being dropped from a conveyer belt at the rate of $\mathrm{M} \mathrm{kg} / \mathrm{s}$. The force necessary to keep the belt moving with a constant velocity of $\mathrm{v} \mathrm{m} / \mathrm{s}$ will be
1 2 Mv newton
2 Mv newton
3 $\frac{M v}{2}$ newton
4 zero
Explanation:
$\begin{aligned} F & =\frac{d p}{d t}=\frac{d}{d t}(m v)=v \frac{d m}{d t} \\ & =v \times M=M v \text { newton } \end{aligned}$
160800
A ball of mass $25 \mathrm{~g}$, moving with a velocity of $2 \mathrm{~ms}^{-1}$ is stopped within $5 \mathrm{~cm}$, The average resistance offered to the ball is
160801
A batsman deflects a ball by an angle of $90^{\circ}$ without changing its initial speed which is equal to $54 \sqrt{2} \mathbf{~ k m h}^{-1}$. Mass of the ball is $0.15 \mathrm{~kg}$. What is the impulse imparted to the ball?
1 $4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction opposite the initial velocity
2 $4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction of the initial velocity
3 $4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction along the bisector of the initial and final directions of the velocity
4 $45 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction of the final velocity
Explanation:
$4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction along the bisector of the initial and final directions of the velocity
NCERT-I-96
2 RBTS PAPER
160802
A man of mass $70 \mathrm{~kg}$ stands on a weighing scale in a lift which is moving upwards with a uniform speed of $10 \mathrm{~ms}^{-1}$, what would be the reading on the scale?
1 $750 \mathrm{~N}$
2 $700 \mathrm{~N}$
3 $350 \mathrm{~N}$
4 $1050 \mathrm{~N}$
Explanation:
Given that mass of the man, $\mathrm{m}$ is $=70 \mathrm{~kg}$ Acceleration, $\mathrm{a}=0$ Using Newton's second law of motion, we can write the equation of motion as: $\mathrm{R}-\mathrm{mg}=\mathrm{ma}$ Where ma is the net force acting on the man. As the lift is moving at a uniform speed, so the acceleration $\mathrm{a}=0$ $\therefore \quad R=m g=70 \times 10=700 N$
NLI Expert
2 RBTS PAPER
160803
Sand is being dropped from a conveyer belt at the rate of $\mathrm{M} \mathrm{kg} / \mathrm{s}$. The force necessary to keep the belt moving with a constant velocity of $\mathrm{v} \mathrm{m} / \mathrm{s}$ will be
1 2 Mv newton
2 Mv newton
3 $\frac{M v}{2}$ newton
4 zero
Explanation:
$\begin{aligned} F & =\frac{d p}{d t}=\frac{d}{d t}(m v)=v \frac{d m}{d t} \\ & =v \times M=M v \text { newton } \end{aligned}$
160800
A ball of mass $25 \mathrm{~g}$, moving with a velocity of $2 \mathrm{~ms}^{-1}$ is stopped within $5 \mathrm{~cm}$, The average resistance offered to the ball is
160801
A batsman deflects a ball by an angle of $90^{\circ}$ without changing its initial speed which is equal to $54 \sqrt{2} \mathbf{~ k m h}^{-1}$. Mass of the ball is $0.15 \mathrm{~kg}$. What is the impulse imparted to the ball?
1 $4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction opposite the initial velocity
2 $4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction of the initial velocity
3 $4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction along the bisector of the initial and final directions of the velocity
4 $45 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction of the final velocity
Explanation:
$4.5 \mathrm{~kg} \mathrm{~ms}^{-1}$ in the direction along the bisector of the initial and final directions of the velocity
NCERT-I-96
2 RBTS PAPER
160802
A man of mass $70 \mathrm{~kg}$ stands on a weighing scale in a lift which is moving upwards with a uniform speed of $10 \mathrm{~ms}^{-1}$, what would be the reading on the scale?
1 $750 \mathrm{~N}$
2 $700 \mathrm{~N}$
3 $350 \mathrm{~N}$
4 $1050 \mathrm{~N}$
Explanation:
Given that mass of the man, $\mathrm{m}$ is $=70 \mathrm{~kg}$ Acceleration, $\mathrm{a}=0$ Using Newton's second law of motion, we can write the equation of motion as: $\mathrm{R}-\mathrm{mg}=\mathrm{ma}$ Where ma is the net force acting on the man. As the lift is moving at a uniform speed, so the acceleration $\mathrm{a}=0$ $\therefore \quad R=m g=70 \times 10=700 N$
NLI Expert
2 RBTS PAPER
160803
Sand is being dropped from a conveyer belt at the rate of $\mathrm{M} \mathrm{kg} / \mathrm{s}$. The force necessary to keep the belt moving with a constant velocity of $\mathrm{v} \mathrm{m} / \mathrm{s}$ will be
1 2 Mv newton
2 Mv newton
3 $\frac{M v}{2}$ newton
4 zero
Explanation:
$\begin{aligned} F & =\frac{d p}{d t}=\frac{d}{d t}(m v)=v \frac{d m}{d t} \\ & =v \times M=M v \text { newton } \end{aligned}$
