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Question 1 |
The $9\ kg$ block shown below is at rest.
The size of angle $θ$ is,
56.5° | |
58.0° | |
67.2° | |
63.4° |
Question 1 Explanation:
The Free Body Diagram of the block is,
Sum of Vertical Forces
$= T_1 \sinθ-mg $ $ = 0 → T_1 \sinθ = mg$
Sum of Horizontal Forces
$= T_1 \cosθ-T_2 $ $ = 0 → T_1 \cosθ = T_2$
$\tanθ = \dfrac{mg}{T_2}$
Using $m = 9\ kg, T_2 = 45\ N $ and $g = 10\ m/s^2,$
$\tanθ = 2$
$→ θ = 63.4°$
Sum of Vertical Forces
$= T_1 \sinθ-mg $ $ = 0 → T_1 \sinθ = mg$
Sum of Horizontal Forces
$= T_1 \cosθ-T_2 $ $ = 0 → T_1 \cosθ = T_2$
$\tanθ = \dfrac{mg}{T_2}$
Using $m = 9\ kg, T_2 = 45\ N $ and $g = 10\ m/s^2,$
$\tanθ = 2$
$→ θ = 63.4°$
Question 2 |
Two blocks are at rest over a rough surface whose coefficient of friction $μ_s$ is $0.3$.
The minimum force $F$ needed to set the system of blocks in motion is,
$30\ N$ | |
$33\ N$ | |
$36\ N$ | |
$39\ N$ |
Question 2 Explanation:
The Free Body Diagram of the system of blocks is,
Sum of vertical forces
$= N_1- mg = 0 → N_1 = mg$
Sum of horizontal forces
$= F- N-μ_s mg $ $ = 0 → F $ $ = N+μ_s mg …(1)$ And,
Sum of vertical forces
$= N_2-Mg = 0 → N_2 = Mg$
Sum of horizontal forces
$= N-μ_s Mg $ $ = 0 → N = μ_s Mg …(2)$
Using (1) and (2),
$F = μ_s g(M+m)$
Using $m = 4\ kg, M = 8\ kg, μ_s = 0.3$ and $g = 10\ m/s^2$
$F = 0.3×10×(4+8)$
$F = 36\ N$
Sum of vertical forces
$= N_1- mg = 0 → N_1 = mg$
Sum of horizontal forces
$= F- N-μ_s mg $ $ = 0 → F $ $ = N+μ_s mg …(1)$ And,
Sum of vertical forces
$= N_2-Mg = 0 → N_2 = Mg$
Sum of horizontal forces
$= N-μ_s Mg $ $ = 0 → N = μ_s Mg …(2)$
Using (1) and (2),
$F = μ_s g(M+m)$
Using $m = 4\ kg, M = 8\ kg, μ_s = 0.3$ and $g = 10\ m/s^2$
$F = 0.3×10×(4+8)$
$F = 36\ N$
Question 3 |
Calculate the average force acting on a block of mass $6.2\ kg$ if it accelerates from rest to $28\ m/s$ in $5.2 \ s$.
$38.2\ N$ | |
$45.3\ N$ | |
$28.8\ N$ | |
$33.4\ N$ |
Question 3 Explanation:
Average acceleration $= \dfrac{28-0}{5.2} = 5.4\ m/s^2$
Average force $= 6.2×5.4 = 33.4\ N$
Average force $= 6.2×5.4 = 33.4\ N$
Question 4 |
A ball of mass $2\ kg$ is initially at rest at the top of a smooth spherical surface of radius $5\ cm$. It is gently pushed so that it starts moving. After rolling through an angle $θ$ it loses contact with the surface.
What is the contact force when the ball loses contact with the surface?
$0\ N$, because the two surfaces are no longer in contact | |
$1.6\ N$, because the contact force is causing centripetal acceleration | |
$0\ N$, because the gravity can provide sufficient centripetal force | |
Cannot be determined, because the nature of the surfaces is not known |
Question 4 Explanation:
Contact force in this case is the normal force and when the two surfaces are no longer in contact the normal force must be zero.
Question 5 |
An object of mass $2\ kg$ starts sliding on a rough incline. The coefficient of kinetic friction between the mass and the incline is $0.22$. The incline is $10\ m$ long. How much less time does the object take to reach the bottom of the incline if there is no friction? (Take the angle of the incline with the horizontal as $30°$)
$0.47\ s$ | |
$0.54\ s$ | |
$0.11\ s$ | |
$0.27\ s$ |
Question 5 Explanation:
Free body diagram of the mass is,
Sum of forces along the direction of $N$
$= N-mg \cosθ $ $ = 0 → N = mg \cosθ$
Sum of forces along the direction of $a$
$mg \sinθ-μN $ $ = ma → mg \sinθ-μmg \cosθ $ $ = ma$
$g(\sinθ-μ \cosθ ) = a$
Using $θ = 30°, g = 10\ m/s^2 $ and $ μ = 0.22$
For rough incline, $a = 3.09\ m/s^2$
For smooth incline, $a = 5.00\ m/s^2$
$s = \dfrac{1}{2} at^2$ for an object starting from rest $(\text{Here}\ s = 10\ m)$
Time taken to reach the bottom of the rough incline
$= \sqrt{\dfrac{2s}{a}} = 2.54\ s$
Time taken to reach the bottom of the smooth incline
$\sqrt{\dfrac{2s}{a}} = 2\ s$
Time difference $= 2.54-2 = 0.54\ s$
Sum of forces along the direction of $N$
$= N-mg \cosθ $ $ = 0 → N = mg \cosθ$
Sum of forces along the direction of $a$
$mg \sinθ-μN $ $ = ma → mg \sinθ-μmg \cosθ $ $ = ma$
$g(\sinθ-μ \cosθ ) = a$
Using $θ = 30°, g = 10\ m/s^2 $ and $ μ = 0.22$
For rough incline, $a = 3.09\ m/s^2$
For smooth incline, $a = 5.00\ m/s^2$
$s = \dfrac{1}{2} at^2$ for an object starting from rest $(\text{Here}\ s = 10\ m)$
Time taken to reach the bottom of the rough incline
$= \sqrt{\dfrac{2s}{a}} = 2.54\ s$
Time taken to reach the bottom of the smooth incline
$\sqrt{\dfrac{2s}{a}} = 2\ s$
Time difference $= 2.54-2 = 0.54\ s$
Question 6 |
Look at the below spring-mass system. The springs are identical in length and nature.
What is the net acceleration of the mass when it is displaced slightly from the rest position by $x$?
$\dfrac{2kx}{m}$ | |
$\dfrac{kx}{m}$ | |
$0$ | |
$\dfrac{4kx}{m}$ |
Question 6 Explanation:
Sum of horizontal forces $= kx+kx = ma$
$2kx = ma$
$a = \dfrac{2kx}{m}$
$2kx = ma$
$a = \dfrac{2kx}{m}$
Question 7 |
What is the apparent weight of a block of mass $m$, going up in an elevator with an acceleration $a$?
$m(g-2a)$ | |
$2m(g-a)$ | |
$m(g+a)$ | |
$\dfrac{1}{2} ma$ |
Question 7 Explanation:
Sum of vertical forces $= N-mg = ma$
$N = m(g+a)$
An object feels its weight due to the normal force acting from the other object.
Thus, apparent weight $= m(g+a)$
$N = m(g+a)$
An object feels its weight due to the normal force acting from the other object.
Thus, apparent weight $= m(g+a)$
Question 8 |
Questions 10, 11 and 12 are based on the below information.
Two blocks, kept on a smooth incline plane, are connected by a massless string going over a pulley, as shown below.
If the two blocks are at rest, then which of the below relations must be correct?
$\cotθ = 0.5$ | |
$\tanθ = 0.5$ | |
$\secθ = 2.5$ | |
$\tanθ = \sqrt2$ |
Question 8 Explanation:
Sum of forces along the incline on $2m$
$= T-2mg\; \sinθ = 0 → T $ $ = 2mg\; \sinθ$
Sum of forces along the incline on $m$
$= T-mg\; \sin(90-θ) $ $ = 0 → T = mg\; \cosθ$
$→ 2 \tanθ = 1\ $ OR $\ \tanθ = 0.5$
$= T-2mg\; \sinθ = 0 → T $ $ = 2mg\; \sinθ$
Sum of forces along the incline on $m$
$= T-mg\; \sin(90-θ) $ $ = 0 → T = mg\; \cosθ$
$→ 2 \tanθ = 1\ $ OR $\ \tanθ = 0.5$
Question 9 |
Two blocks, kept on a smooth incline plane, are connected by a massless string going over a pulley, as shown below.
If the actual $θ = 20°$, then what will be the acceleration of the lighter block?
$0.08g,$ $\text{down the incline}$ | |
$0.12g,$ $\text{up the incline}$ | |
$0.18g,$ $\text{up the incline}$ | |
$0.15g,$ $\text{down the incline}$ |
Question 9 Explanation:
{Assume a certain direction of acceleration for calculation if answer is negative the actual direction will be opposite to the assumption.}
Sum of forces along the incline on $m$
$= mg\; \cosθ-T = ma$
(acceleration down the incline)
Sum of forces along the incline on $2m$
$= T-2mg\; \sinθ = 2ma$
(acceleration up the incline)
Adding the two equations,
$mg \cosθ-2mg\; \sinθ = 3ma$
$mg(\cosθ-2 \sinθ ) = 3ma$
Using $θ = 20°$,
$a = \dfrac{g}{3} (0.25)$
$a = 0.08g$
Sum of forces along the incline on $m$
$= mg\; \cosθ-T = ma$
(acceleration down the incline)
Sum of forces along the incline on $2m$
$= T-2mg\; \sinθ = 2ma$
(acceleration up the incline)
Adding the two equations,
$mg \cosθ-2mg\; \sinθ = 3ma$
$mg(\cosθ-2 \sinθ ) = 3ma$
Using $θ = 20°$,
$a = \dfrac{g}{3} (0.25)$
$a = 0.08g$
Question 10 |
Two blocks, kept on a smooth incline plane, are connected by a massless string going over a pulley, as shown below.
Why does the pulley stay in the same place?
The friction force between the pulley and the incline balances the tension forces | |
The gravitational force of the pulley balances the tension forces | |
The normal reaction between the pulley and the incline balances the tension forces | |
Tension forces are of equal strength and so no other force is needed to balance them |
Question 10 Explanation:
The pulley is acted upon by tension forces in different directions.
Thus, the contact (Normal) force balances the tension forces acting on the pulley.
Thus, the contact (Normal) force balances the tension forces acting on the pulley.
Question 11 |
A marble of mass $m_1$, tied to a string, performs uniform circular motion.
A block of mass $m_2$ is attached to the other end of the string. The velocity of the marble at which the tension force of the string is enough to balance the weight of the block is,
$\sqrt{2gr}$ | |
$\sqrt{\dfrac{2m_2 gr}{(m_1+m_2)}}$ | |
$\sqrt{\dfrac{m_1 gr}{m_2}}$ | |
$\sqrt{\dfrac{m_2 gr}{m_1}}$ |
Question 11 Explanation:
For Marble,
Tension $= T = \dfrac{m_1 v^2}{r}$
For Block,
Tension $= T = m_2 g$
Thus, $\dfrac{m_1 v^2}{r} = m_2 g → v^2 $ $ = \dfrac{m_2 gr}{m_1}$
Or $v = \sqrt{\dfrac{m_2 gr}{m_1}}$
Tension $= T = \dfrac{m_1 v^2}{r}$
For Block,
Tension $= T = m_2 g$
Thus, $\dfrac{m_1 v^2}{r} = m_2 g → v^2 $ $ = \dfrac{m_2 gr}{m_1}$
Or $v = \sqrt{\dfrac{m_2 gr}{m_1}}$
Question 12 |
Find the value of gravitational acceleration at a height of $2R_e$ from the Earth’s surface as a percentage of gravitational acceleration at the surface. (Here, $R_e$ is the radius of the Earth)
18.3% | |
14.2% | |
11.1% | |
9.9% |
Question 12 Explanation:
$g = \dfrac{GM}{R_e^2}$
While,
$g_h = \dfrac{GM}{(R_e+2R_e )^2} $ $ = \dfrac{GM}{9R_e^2}$
$\dfrac{g_h}{g} = \dfrac{GM}{9R_e^2}÷\dfrac{GM}{R_e^2} $ $ = \dfrac{1}{9}$
Thus, $\dfrac{g_h}{g} = 11.1 \% $
While,
$g_h = \dfrac{GM}{(R_e+2R_e )^2} $ $ = \dfrac{GM}{9R_e^2}$
$\dfrac{g_h}{g} = \dfrac{GM}{9R_e^2}÷\dfrac{GM}{R_e^2} $ $ = \dfrac{1}{9}$
Thus, $\dfrac{g_h}{g} = 11.1 \% $
Question 13 |
Andrew draws a schematic diagram for astronomical objects.
What is the net gravitational force acting on the heavier object?
$\dfrac{10\sqrt{2} Gm^2}{y^2}$ $\text{along the diagonal}$ | |
$\dfrac{10 Gm^2}{y^2}$ $\text{towards the left}$ | |
$\dfrac{10\sqrt{3} Gm^2}{y^2}$ $\text{along the diagonal}$ | |
$\dfrac{10 Gm^2}{y^2}$ $\text{coming out of the plane}$ |
Question 13 Explanation:
Force of gravity on $10\ m$ is equal to the vector sum of the forces from the left and bottom by $m$ and $m$
Force from left $F_l = \dfrac{10Gm^2}{y^2}$
Force from bottom $F_b = \dfrac{10Gm^2}{y^2}$
Net force = $\sqrt{F_l^2+F_b^2} = \dfrac{10\sqrt{2} Gm^2}{y^2}$
(along the diagonal)
Force from left $F_l = \dfrac{10Gm^2}{y^2}$
Force from bottom $F_b = \dfrac{10Gm^2}{y^2}$
Net force = $\sqrt{F_l^2+F_b^2} = \dfrac{10\sqrt{2} Gm^2}{y^2}$
(along the diagonal)
Question 14 |
What is the apparent weight of an object inside a satellite?
Same as the actual weight | |
2 times the actual weight | |
½ times the actual weight | |
0 times the actual weight |
Question 14 Explanation:
An object feels its own weight due to the normal force acting on it.
Since inside a satellite all the objects are in free fall towards the Earth (each is moving with the same acceleration).
Hence, there are no normal forces acting on an object in a satellite.
Thus, the apparent weight is 0.
Since inside a satellite all the objects are in free fall towards the Earth (each is moving with the same acceleration).
Hence, there are no normal forces acting on an object in a satellite.
Thus, the apparent weight is 0.
Question 15 |
A book weighs $28\ N$ on the surface of the Earth. What is the weight of the book on another planet whose radius is twice that of Earth and Mass thrice that of Earth?
$25\ N$ | |
$21\ N$ | |
$32\ N$ | |
$36\ N$ |
Question 15 Explanation:
Mass of the book $= \dfrac{28}{10} = 2.8\ kg$
(Using $g_e = 10\ m/s^2$)
Gravitational acceleration on the planet
$= \dfrac{GM_p}{R_p^2} = \dfrac{3GM_e}{4R_e^2} = 0.75\ g$
Weight on the planet
$= 0.75\ mg $ $ = 0.75×10×2.8 $ $ = 21\ N$
(Using $g_e = 10\ m/s^2$)
Gravitational acceleration on the planet
$= \dfrac{GM_p}{R_p^2} = \dfrac{3GM_e}{4R_e^2} = 0.75\ g$
Weight on the planet
$= 0.75\ mg $ $ = 0.75×10×2.8 $ $ = 21\ N$
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