NEET Test Series from KOTA - 10 Papers In MS WORD
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2 RBTS PAPER
160774
Two particles of equal masses are revolving in circular paths of radii $r_1$ and $r_2$ respectively with the same speed. The ratio of their centripetal force is:
160775
A rope of length $6 \mathrm{~m}$ is kept on a frictionless surface and a force of $18 \mathrm{~N}$ is applied to one of its end. Find tension in the rope at the mid point.
1 $1 \mathrm{~N}$
2 $3 \mathrm{~N}$
3 $9 \mathrm{~N}$
4 $6 \mathrm{~N}$
Explanation:
A rope of length $L$ and mass $M$ is pulled by a constant force $\mathrm{F}$, then tension in rope of any point C. Tension in rope, $T=\frac{M}{L}(L-x) \times \frac{F}{M}$ $\begin{aligned} T & =\frac{F(L-x)}{L} \\ & =\frac{18(6-3)}{6}=9 \mathrm{~N} \end{aligned}$
NLI Expert
2 RBTS PAPER
160776
Which one of the following cannot be explained on the basis of Newton's third law of motion?
1 Rebounding of a ball from a wall
2 Motion of jet in the sky
3 Returning back of body, thrown above
4 Rowing of a boat in a pond
Explanation:
Returning back of a body thrown upwards ighot a reaction.
NCERT-I-97
2 RBTS PAPER
160777
A gun fires a bullet of mass $\mathbf{5 0} \mathrm{g}$ with a velocity of $30 \mathrm{~ms}^{-1}$. Because of this, the gun is pushed back with a velocity of $1 \mathrm{~ms}^{-1}$. The mass of the gun is
1 $5.5 \mathrm{~kg}$
2 $1.5 \mathrm{~kg}$
3 $3.5 \mathrm{~kg}$
4 $0.5 \mathrm{~kg}$
Explanation:
By conservation of momentum $M V=m v$ $\mathrm{M}=\frac{\mathrm{mv}}{\mathrm{V}}=\frac{0.05 \times 30}{1}=15 \mathrm{~kg}$
160774
Two particles of equal masses are revolving in circular paths of radii $r_1$ and $r_2$ respectively with the same speed. The ratio of their centripetal force is:
160775
A rope of length $6 \mathrm{~m}$ is kept on a frictionless surface and a force of $18 \mathrm{~N}$ is applied to one of its end. Find tension in the rope at the mid point.
1 $1 \mathrm{~N}$
2 $3 \mathrm{~N}$
3 $9 \mathrm{~N}$
4 $6 \mathrm{~N}$
Explanation:
A rope of length $L$ and mass $M$ is pulled by a constant force $\mathrm{F}$, then tension in rope of any point C. Tension in rope, $T=\frac{M}{L}(L-x) \times \frac{F}{M}$ $\begin{aligned} T & =\frac{F(L-x)}{L} \\ & =\frac{18(6-3)}{6}=9 \mathrm{~N} \end{aligned}$
NLI Expert
2 RBTS PAPER
160776
Which one of the following cannot be explained on the basis of Newton's third law of motion?
1 Rebounding of a ball from a wall
2 Motion of jet in the sky
3 Returning back of body, thrown above
4 Rowing of a boat in a pond
Explanation:
Returning back of a body thrown upwards ighot a reaction.
NCERT-I-97
2 RBTS PAPER
160777
A gun fires a bullet of mass $\mathbf{5 0} \mathrm{g}$ with a velocity of $30 \mathrm{~ms}^{-1}$. Because of this, the gun is pushed back with a velocity of $1 \mathrm{~ms}^{-1}$. The mass of the gun is
1 $5.5 \mathrm{~kg}$
2 $1.5 \mathrm{~kg}$
3 $3.5 \mathrm{~kg}$
4 $0.5 \mathrm{~kg}$
Explanation:
By conservation of momentum $M V=m v$ $\mathrm{M}=\frac{\mathrm{mv}}{\mathrm{V}}=\frac{0.05 \times 30}{1}=15 \mathrm{~kg}$
160774
Two particles of equal masses are revolving in circular paths of radii $r_1$ and $r_2$ respectively with the same speed. The ratio of their centripetal force is:
160775
A rope of length $6 \mathrm{~m}$ is kept on a frictionless surface and a force of $18 \mathrm{~N}$ is applied to one of its end. Find tension in the rope at the mid point.
1 $1 \mathrm{~N}$
2 $3 \mathrm{~N}$
3 $9 \mathrm{~N}$
4 $6 \mathrm{~N}$
Explanation:
A rope of length $L$ and mass $M$ is pulled by a constant force $\mathrm{F}$, then tension in rope of any point C. Tension in rope, $T=\frac{M}{L}(L-x) \times \frac{F}{M}$ $\begin{aligned} T & =\frac{F(L-x)}{L} \\ & =\frac{18(6-3)}{6}=9 \mathrm{~N} \end{aligned}$
NLI Expert
2 RBTS PAPER
160776
Which one of the following cannot be explained on the basis of Newton's third law of motion?
1 Rebounding of a ball from a wall
2 Motion of jet in the sky
3 Returning back of body, thrown above
4 Rowing of a boat in a pond
Explanation:
Returning back of a body thrown upwards ighot a reaction.
NCERT-I-97
2 RBTS PAPER
160777
A gun fires a bullet of mass $\mathbf{5 0} \mathrm{g}$ with a velocity of $30 \mathrm{~ms}^{-1}$. Because of this, the gun is pushed back with a velocity of $1 \mathrm{~ms}^{-1}$. The mass of the gun is
1 $5.5 \mathrm{~kg}$
2 $1.5 \mathrm{~kg}$
3 $3.5 \mathrm{~kg}$
4 $0.5 \mathrm{~kg}$
Explanation:
By conservation of momentum $M V=m v$ $\mathrm{M}=\frac{\mathrm{mv}}{\mathrm{V}}=\frac{0.05 \times 30}{1}=15 \mathrm{~kg}$
160774
Two particles of equal masses are revolving in circular paths of radii $r_1$ and $r_2$ respectively with the same speed. The ratio of their centripetal force is:
160775
A rope of length $6 \mathrm{~m}$ is kept on a frictionless surface and a force of $18 \mathrm{~N}$ is applied to one of its end. Find tension in the rope at the mid point.
1 $1 \mathrm{~N}$
2 $3 \mathrm{~N}$
3 $9 \mathrm{~N}$
4 $6 \mathrm{~N}$
Explanation:
A rope of length $L$ and mass $M$ is pulled by a constant force $\mathrm{F}$, then tension in rope of any point C. Tension in rope, $T=\frac{M}{L}(L-x) \times \frac{F}{M}$ $\begin{aligned} T & =\frac{F(L-x)}{L} \\ & =\frac{18(6-3)}{6}=9 \mathrm{~N} \end{aligned}$
NLI Expert
2 RBTS PAPER
160776
Which one of the following cannot be explained on the basis of Newton's third law of motion?
1 Rebounding of a ball from a wall
2 Motion of jet in the sky
3 Returning back of body, thrown above
4 Rowing of a boat in a pond
Explanation:
Returning back of a body thrown upwards ighot a reaction.
NCERT-I-97
2 RBTS PAPER
160777
A gun fires a bullet of mass $\mathbf{5 0} \mathrm{g}$ with a velocity of $30 \mathrm{~ms}^{-1}$. Because of this, the gun is pushed back with a velocity of $1 \mathrm{~ms}^{-1}$. The mass of the gun is
1 $5.5 \mathrm{~kg}$
2 $1.5 \mathrm{~kg}$
3 $3.5 \mathrm{~kg}$
4 $0.5 \mathrm{~kg}$
Explanation:
By conservation of momentum $M V=m v$ $\mathrm{M}=\frac{\mathrm{mv}}{\mathrm{V}}=\frac{0.05 \times 30}{1}=15 \mathrm{~kg}$
