iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 24th June Morning Shift
A particle experiences a variable force $\overrightarrow F = \left( {4x\widehat i + 3{y^2}\widehat j} \right)$ in a horizontal x-y plane. Assume distance in meters and force is newton. If the particle moves from point (1, 2) to point (2, 3) in the x-y plane, then Kinetic Energy changes by :
A.
50.0 J
B.
12.5 J
C.
25.0 J
D.
0 J
Correct Answer: C
Explanation:
$W = \int {\overrightarrow F \,.\,d\overrightarrow r } $
The value of the work done by the air friction is $-$ 0.68 mgh.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 31st August Evening Shift
A block moving horizontally on a smooth surface with a speed of 40 m/s splits into two parts with masses in the ratio of 1 : 2. If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is :-
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th July Evening Shift
Given below is the plot of a potential energy function U(x) for a system, in which a particle is in one dimensional motion, while a conservative force F(x) acts on it. Suppose that Emech = 8 J, the incorrect statement for this system is :
[ where K.E. = kinetic energy ]
A.
at x > x4 K.E. is constant throughout the region.
B.
at x < x1, K.E. is smallest and the particle is moving at the slowest speed.
C.
at x = x2, K.E. is greatest and the particle is moving at the fastest speed.
D.
at x = x3, K.E. = 4 J.
Correct Answer: B
Explanation:
Emech. = 8J
(A) at x > x4
U = constant = 6J
K = Emech. $-$ U = 2J = constant
(B) at x < x1
U = constant = 8J
K = Emech $-$ U = 8 $-$ 8 - 0 J
Particle is at rest.
(C) At x = x2,
U = 0 $\Rightarrow$ Emech. = K = 8 J
KE is greatest, and particle is moving at fastest speed.
(D) At x = x3
U = 4J
U + K = 8J
K = 4J
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th July Evening Shift
An automobile of mass 'm' accelerates starting from origin and initially at rest, while the engine supplies constant power P. The position is given as a function of time by :
$ \Rightarrow x = {\left( {{{8P} \over {9m}}} \right)^{1/2}}{t^{3/2}}$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 22th July Evening Shift
A porter lifts a heavy suitcase of mass 80 kg and at the destination lowers it down by a distance of 80 cm with a constant velocity. Calculate the work done by the porter in lowering the suitcase.
(take g = 9.8 ms$-$2)
A.
+627.2 J
B.
$-$62720.0 J
C.
$-$627.2 J
D.
784.0 J
Correct Answer: C
Explanation:
$W = - N \times \Delta x$
$ = - 80 \times 9.8 \times {{80} \over {100}}$
$ = - 627.2$ J
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 20th July Evening Shift
A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time 't' is proportional to :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 18th March Morning Shift
A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time 't' is proportional to :-
A.
t2/3
B.
t3/2
C.
t
D.
t1/2
Correct Answer: B
Explanation:
$P = F.v = mav$
$P = {{mvdv} \over {dt}}$
$\int\limits_0^t {Pdt} = m\int\limits_0^v {vdv} $
$Pt = {{m{v^2}} \over 2}$
$v = \sqrt {{{2Pt} \over m}} $
${{dx} \over {dt}} = \sqrt {{{2Pt} \over m}} $
$\int {dx} = \int {\sqrt {{{2Pt} \over m}} } dt$
$x \propto {t^{3/2}}$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 17th March Morning Shift
A boy is rolling a 0.5 kg ball on the frictionless floor with the speed of 20 ms-1. The ball gets deflected by an obstacle on the way. After deflection it moves with 5% of its initial kinetic energy. What is the speed of the ball now?
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 6th September Morning Slot
If the potential energy between two molecules is given by
U = $ - {A \over {{r^6}}} + {B \over {{r^{12}}}}$,
then at equilibrium,
separation between molecules, and the potential energy are :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 4th September Evening Slot
A person pushes a box on a rough horizontal plateform surface. He applies a force of 200 N over a
distance of 15 m. Thereafter, he gets progressively tired and his applied force reduces linearly with
distance to 100 N. The total distance through which the box has been moved is 30 m. What is the
work done by the person during the total movement of the box?
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 3rd September Evening Slot
A particle is moving unidirectionally on a horizontal plane under the action of a constant power
supplying energy source. The displacement (s) - time (t) graph that describes the motion of the
particle is (graphs are drawn schematically and are not to scale) :
A.
B.
C.
D.
Correct Answer: B
Explanation:
$P = Fv$
$P = m{{dv} \over {dt}}v$
$ \Rightarrow $ $vdv = {P \over m}dt$
Integrating both sides, we get
${V^2} = k't$
$ \Rightarrow $ V = k"$\sqrt t $
$ \Rightarrow $ ${{ds} \over {dt}}$ = k"$\sqrt t $
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 9th January Morning Slot
Consider a force $\overrightarrow F = - x\widehat i + y\widehat j$
. The work done
by this force in moving a particle from point
A(1, 0) to B(0, 1) along the line segment is :
(all quantities are in SI units)
A.
2
B.
${1 \over 2}$
C.
1
D.
${3 \over 2}$
Correct Answer: C
Explanation:
W = $\int {\overrightarrow F } .d\overrightarrow r $
= $\int {\left( { - x\widehat i + y\widehat j} \right)} .\left( {dx\widehat i + dy\widehat j} \right)$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 7th January Evening Slot
An elevator in a building can carry a maximum of 10 persons, with the average mass of each
person being 68 kg, The mass of the elevator itself is 920 kg and it moves with a constant speed
of 3 m/s. The frictional force opposing the motion is 6000 N. If the elevator is moving up with its
full capacity, the power delivered by the motor to the elevator (g = 10 m/s2) must be at least :
A.
48000 W
B.
62360 W
C.
56300 W
D.
66000 W
Correct Answer: D
Explanation:
Net force on motor will be
Fm = [920 + 68(10)]g + 6000
= 22000 N
So, required power for motor
P = Fm.V = 22000$ \times $3 = 66000 W
2020
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 7th January Morning Slot
A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to : (1 HP = 746 W, g = 10 ms-2)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 9th April Morning Slot
A uniform cable of mass 'M' and length 'L' is
placed on a horizontal surface such that its (1/n)th
part is hanging below the edge of the
surface. To lift the hanging part of the cable upto
the surface, the work done should be :
A.
${{2MgL} \over {{n^2}}}$
B.
nMgL
C.
${{MgL} \over {2{n^2}}}$
D.
${{MgL} \over {{n^2}}}$
Correct Answer: C
Explanation:
To solve this problem, we need to determine the work done to lift the hanging part of the cable up to the surface. The work done lifting a small element of the cable will be the weight of the element times the distance it has to be lifted.
Let's take a small section of the cable at a depth $x$ below the surface. This section has a length of $dx$, so its mass is $(M/L)dx$ where $(M/L)$ is the linear mass density of the cable.
The work $dW$ done to lift this small section up to the surface is the weight of the section times the distance it has to be lifted :
$dW = (M/L)gdx \times x$.
Integrating this expression from 0 to L/n (the length of the hanging part of the cable) gives the total work done :
$W = \int\limits_{0}^{L/n} (M/L)gxdx$
$= (Mg/L) \int\limits_{0}^{L/n} xdx$
$= (Mg/L) \times [x^2/2]_{0}^{L/n}$
$= (Mg/L) \times [L^2/(2n^2)]$
$= MgL/(2n^2)$
So the work done to lift the hanging part of the cable up to the surface is $MgL/(2n^2)$.
Therefore, the correct answer is Option C :
$\frac{MgL}{2n^2}$
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 8th April Morning Slot
A particle moves in one dimension from rest
under the influence of a force that varies with
the distance travelled by the particle as shown
in the figure. The kinetic energy of the particle
after it has travelled 3m is :
A.
6.5 J
B.
2.5 J
C.
5 J
D.
4 J
Correct Answer: A
Explanation:
According to work energy theorem.
Work done by force on the particle = Change in KE
Work done = Area under F-x graph = $\int {F.dx} $
= $2 \times 2 + {{(2 + 3) \times 1} \over 2}$
W = KEfinal – KEinitial = 6.5
KEinitial = 0
KEfinal = 6.5 J
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 10th January Evening Slot
A particle which is experiencing a force, given by $\overrightarrow F = 3\widehat i - 12\widehat j,$ undergoes a displacement of $\overrightarrow d = 4\overrightarrow i $ particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?
A.
9 J
B.
10 J
C.
12 J
D.
15 J
Correct Answer: D
Explanation:
Work done = $\overrightarrow F \cdot \overrightarrow d $
$=$ 12 J
work energy theorem
wnet $=$ $\Delta $K.E.
12 $=$ Kf $-$ 3
Kf = 15 J
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 10th January Morning Slot
A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in figure. Work done by normal reaction on block in time is -
A.
${{m{g^2}{t^2}} \over 8}$
B.
${{3m{g^2}{t^2}} \over 8}$
C.
$-$ ${{m{g^2}{t^2}} \over 8}$
D.
0
Correct Answer: B
Explanation:
N $-$ mg = ${{mg} \over 2}$ $ \Rightarrow $ N = ${{3mg} \over 2}$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 9th January Evening Slot
A force acts on a 2 kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds ?
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 9th January Morning Slot
A block of mass m, lying on a smooth horizontal surface, is attached to a sring (of negligible mass) of spring constant k. The other end of the spring is fixed, as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force F, the maximum speed of the block is :
A.
${{2F} \over {\sqrt {mk} }}$
B.
${F \over {\pi \sqrt {mk} }}$
C.
${{\pi F} \over {\sqrt {mk} }}$
D.
${F \over {\sqrt {mk} }}$
Correct Answer: D
Explanation:
When block is pulled x distance with F force, then it is distance with F force, then it is at maximum position
$ \therefore $ F = Kx
$ \Rightarrow $ x = ${F \over K}$
Applying work energy theorem, work done by all the forces = change in Kinetic energy.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 16th April Morning Slot
Two particles of the same mass m are moving in circular orbits because of force, given by $F\left( r \right) = {{ - 16} \over r} - {r^3}$
The first particle is at a distance r = 1, and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 16th April Morning Slot
A body of mass m starts moving from rest along x-axis so that its velocity varies as $\upsilon = a\sqrt s $ where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t seconds after the start of the motion is :
A.
${1 \over 8}\,$ m a4 t2
B.
8 m a4 t2
C.
4 m a4 t2
D.
${1 \over 4}\,$ m a4 t2
Correct Answer: A
Explanation:
Given,
$\upsilon $ = a $\sqrt s $
$ \Rightarrow $$\,\,\,$ ${{ds} \over {dt}} = a\sqrt s $
We know kinetic energy of particle, K = ${1 \over 2}m{v^2}$ = ${k \over {2{r^2}}}$
As Total energy = Kinetic energy + Potential energy
So Total energy = ${k \over {2{r^2}}}$ $ - {k \over {2{r^2}}}$ = 0
2017
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Online) 8th April Morning Slot
An object is dropped from a height h from the ground. Every time it hits the ground it looses 50% of its kinetic energy. The total distance covered as t $ \to $ $\infty $ is :
A.
3h
B.
$\infty $
C.
${5 \over 3}$h
D.
${8 \over 3}$h
Correct Answer: A
Explanation:
Let,
Kinetic energy (k) = ${1 \over 2}$ m $\upsilon $2 before it hit the ground.
Total distance travelled from the time it first hits the ground to the next time it hits the ground is = ${h \over 2}$ + ${h \over 2}$ = h
So, this will create a infinite geometric progression with the common ration ${1 \over 2}$.
$\therefore\,\,\,$ Total distance covered
= h (distance travelled by the obhect when first dropped, before it hits the ground)
+ (h + ${h \over 2}$ + ${h \over 4}$ + . . . . . . . .$ \propto $)
= h + ${h \over {1 - {1 \over 2}}}$
= h + 2h
= 3h
2017
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Offline)
A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done
by the force during the first 1 sec. will be:
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Offline)
A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is ${1 \over 8}mv_0^2$, the value of k will be:
A.
10-1 kg m-1 s-1
B.
10-3 kg m-1
C.
10-3 kg s-1
D.
10-4 kg m-1
Correct Answer: D
Explanation:
According to the question, final kinetic energy = ${1 \over 8}mv_0^2$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2016 (Online) 10th April Morning Slot
A particle of mass M is moving in a circle of fixed radius R in such a way that its centripetal acceleration at time t is given by n2 R t2 where n is a constant. The power delivered to the particle by the force acting on it, is :
A.
M n2 R2 t
B.
M n R2 t
C.
M n R2 t2
D.
${1 \over 2}$ M n2 R2 t2
Correct Answer: A
Explanation:
We know,
centripetal acceleration = ${{{V^2}} \over R}$
$ \therefore $ According to question,
${{{V^2}} \over R}$ = ${n^2}R{t^2}$
$ \Rightarrow $ V2 = n2 R2 t2
$ \Rightarrow $ V = nRt
$ \Rightarrow $ ${{dV} \over {dt}}$ = nR
Power (P) = Force (F) $ \times $ Velocity (V)
= M ${{dV} \over {dt}}$(V)
= M (nR) (nRt)
= Mn2R2t
2016
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2016 (Online) 9th April Morning Slot
A car of weight W is on an inclined road that rises by 100 m over a distance of 1 km
and applies a constant frictional force ${W \over 20}$ on the car. While moving uphill on the road at a speed of 10 ms−1, the car needs power P. If it needs power ${p \over 2}$ while moving downhill at speed v then value of $\upsilon $ is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2016 (Offline)
A point particle of mass $m,$ moves long the uniformly rough track $PQR$ as shown in the figure. The coefficient of friction, between the particle and the rough track equals $\mu .$ The particle is released, from rest from the point $P$ and it comes to rest at point $R.$ The energies, lost by the ball, over the parts, $PQ$ and $QR$, of the track, are equal to each other , and no energy is lost when particle changes direction from $PQ$ to $QR$.
The value of the coefficient of friction $\mu $ and the distance $x$ $(=QR),$ are, respectively close to:
A.
$0.29$ and $3.5$ $m$
B.
$0.29$ and $6.5$ $m$
C.
$0.2$ and $6.5$ $m$
D.
$0.2$ and $3.5$ $m$
Correct Answer: A
Explanation:
Using work energy theorem for the motion of the particle,
Loss in $P.E.=$ Work done against friction from $p \to Q$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2016 (Offline)
A person trying to lose weight by burning fat lifts a mass of $10$ $kg$ upto a height of $1$ $m$ $1000$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8 \times {10^7}J$ of energy per $kg$ which is converted to mechanical energy with a $20\% $ efficiency rate. Take $g = 9.8\,m{s^{ - 2}}$ :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2014 (Offline)
When a rubber-band is stretched by a distance $x$, it exerts restoring force of magnitude $F = ax + b{x^2}$ where $a$ and $b$ are constants. The work done in stretching the unstretched rubber-band by $L$ is :
This question has Statement $1$ and Statement $2.$ Of the four choices given after the Statements, choose the one that best describes the two Statements.
If two springs ${S_1}$ and ${S_2}$ of force constants ${k_1}$ and ${k_2}$, respectively, are stretched by the same force, it is found that more work is done on spring ${S_1}$ than on spring ${S_2}$.
STATEMENT 1: If stretched by the same amount work done on ${S_1}$, Work done on ${S_1}$ is more than ${S_2}$
STATEMENT 2: ${k_1} < {k_2}$
A.
Statement 1 is false, Statement 2 is true
B.
Statement 1 is true, Statement 2 is false
C.
Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1
D.
Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1
The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}},$ where $a$ and $b$ are constants and $x$ is the distance between the atoms. If the dissociation energy of the molecule is $D = \left[ {U\left( {x = \infty } \right) - {U_{at\,\,equilibrium}}} \right],\,\,D$ is
A.
${{{b^2}} \over {2a}}$
B.
${{{b^2}} \over {12a}}$
C.
${{{b^2}} \over {4a}}$
D.
${{{b^2}} \over {6a}}$
Correct Answer: C
Explanation:
Given $U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}}$
A $2$ $kg$ block slides on a horizontal floor with a speed of $4m/s.$ It strikes a uncompressed spring, and compress it till the block is motionless. The kinetic friction force is $15N$ and spring constant is $10, 000$ $N/m.$ The spring compresses by
A.
$8.5cm$
B.
$5.5cm$
C.
$2.5cm$
D.
$11.0cm$
Correct Answer: B
Explanation:
Let the block compress the spring by $x$ before coming to rest.
Initial kinetic energy of the block $=$ (potential energy of compressed spring) $+$ work done due to friction.
At the highest point the velocity is $u$ $\cos \,{60^ \circ }$ (only the horizontal component remains, the vertical component being zero at the top-most point).
The potential energy of a $1$ $kg$ particle free to move along the $x$-axis is given by $V\left( x \right) = \left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right)J$.
The total mechanical energy of the particle is $2J.$ Then, the maximum speed (in $m/s$) is
A.
${3 \over {\sqrt 2 }}$
B.
${\sqrt 2 }$
C.
${1 \over {\sqrt 2 }}$
D.
$2$
Correct Answer: A
Explanation:
Velocity is maximum when kinetic energy is maximum and when kinetic energy is maximum then potential energy should be minimum
For minimum potential energy,
${{dV} \over {dx}} = 0 $
$\Rightarrow {x^3} - x = 0 $
$\Rightarrow x = \pm 1$
$ \Rightarrow$ Min. Potential energy (P.E.) =$ {1 \over 4} - {1 \over 2} = - {1 \over 4}J$
A ball of mass $0.2$ $kg$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2$ $m$ while applying the force and the ball goes upto $2$ $m$ height further, find the magnitude of the force. (consider $g = 10\,m/{s^2}$).
A.
$4N$
B.
$16$ $N$
C.
$20$ $N$
D.
$22$ $N$
Correct Answer: D
Explanation:
According to energy conservation law,
Work done by the hand and due to gravity = total change in the kinetic energy
Initially the the ball is at rest and finally at top its velocity become zero so total change in kinetic energy $\Delta K$ = 0
${W_{hand}} + {W_{gravity}} = \Delta K$
[Here distance covered would be 0.2 meter for force by hand as force is applied while ball is in contact with hand.
And gravity will still work while ball is in contact with hand so total distance due to gravity would be 2 + 0.2 = 2.2 meter.]
$ \Rightarrow F\left( {0.2} \right) - \left( {0.2} \right)\left( {10} \right)\left( {2.2} \right)$ $ = 0 \Rightarrow F = 22\,N$
A particle of mass $100g$ is thrown vertically upwards with a speed of $5$ $m/s$. The work done by the force of gravity during the time the particle goes up is
A.
$-0.5J$
B.
$-1.25J$
C.
$1.25J$
D.
$0.5J$
Correct Answer: B
Explanation:
Kinetic energy at point of throwing is converted into potential energy of the particle during rise.
A mass of $M$ $kg$ is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of ${45^ \circ }$ with the initial vertical direction is
A.
$Mg\left( {\sqrt 2 + 1} \right)$
B.
$Mg\sqrt 2 $
C.
${{Mg} \over {\sqrt 2 }}$
D.
$Mg\left( {\sqrt 2 - 1} \right)$
Correct Answer: D
Explanation:
From work energy theorem we can say,
Work done by tension $+$ work done by force (applied) $+$ Work done by gravitational force $=$ change in kinetic energy
Here Work done by tension is zero
$ \Rightarrow 0 + F \times AB - Mg \times AC = 0$
$ \Rightarrow F = Mg\left( {{{AC} \over {AB}}} \right) = Mg\left[ {{{1 - {1 \over {\sqrt 2 }}} \over {{1 \over 2}}}} \right]$
[ as $AB = \ell \sin {45^ \circ } = {\ell \over {\sqrt 2 }}$
and $AC = OC - OA = \ell - \ell \,\cos \,{45^ \circ } = \ell \left( {1 - {1 \over {\sqrt 2 }}} \right)$
where $\ell = $ length of the string. ]
$ \Rightarrow F = Mg\left( {\sqrt 2 - 1} \right)$
A bullet fired into a fixed target loses half of its velocity after penetrating $3$ $cm.$ How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?
A.
$2.0$ $cm$
B.
$3.0$ $cm$
C.
$1.0$ $cm$
D.
$1.5$ $cm$
Correct Answer: C
Explanation:
Let $K$ be the initial kinetic energy and $F$ be the resistive force. Then according to work-energy theorem,
$$W = \Delta K$$
A spherical ball of mass $20$ $kg$ is stationary at the top of a hill of height $100$ $m$. It rolls down a smooth surface to the ground, then climbs up another hill of height $30$ $m$ and finally rolls down to a horizontal base at a height of $20$ $m$ above the ground. The velocity attained by the ball is
A.
$20$ $m/s$
B.
$40$ $m/s$
C.
$10\sqrt {30} \,\,\,m/s$
D.
$10\,\,m/s$
Correct Answer: B
Explanation:
Loss in potential energy $=$ gain in kinetic energy
A body of mass $m$ is accelerated uniformly from rest to a speed $v$ in a time $T.$ The instantaneous power delivered to the body as a function of time is given by
A.
${{m{v^2}} \over {{T^2}}}.{t^2}$
B.
${{m{v^2}} \over {{T^2}}}.t$
C.
${1 \over 2}{{m{v^2}} \over {{T^2}}}.{t^2}$
D.
${1 \over 2}{{m{v^2}} \over {{T^2}}}.t$
Correct Answer: B
Explanation:
$u = 0;v = u + aT;v = aT$
Instantaneous power $ = F \times v = m.\,a.\,at = m.{a^2}.t$
$\therefore$ Instantaneous power $ = {{m{v^2}t} \over {{T^2}}}$
The upper half of an inclined plane with inclination $\phi $ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by
A.
$2\,\cos \,\,\phi $
B.
$2\,sin\,\,\phi $
C.
$\,\tan \,\,\phi $
D.
$2\,\tan \,\,\phi $
Correct Answer: D
Explanation:
Let the length of the inclined plane is = $l$. So only ${l \over 2}$ part will have friction.
According to work-energy theorem, $W = \Delta k = 0$
(Since initial and final speeds are zero)
$\therefore$ Work done by friction + Work done by gravity $=0$
A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to
A.
$x$
B.
${e^x}$
C.
${x^2}$
D.
${\log _e}x$
Correct Answer: C
Explanation:
Given that, retardation $ \propto $ displacement
$ \Rightarrow $ $a=-kx$
But we know $a = v{{dv} \over {dx}}\,\,\,\,\,\,\,\,\,$
$\therefore$ ${{vdv} \over {dx}} = - kx $
$\Rightarrow \int\limits_{{v_1}}^{{v_2}} v \,dv = - k\int\limits_0^x {xdx} $
A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particles takes place in a plane. It follows that
A.
its kinetic energy is constant
B.
is acceleration is constant
C.
its velocity is constant
D.
it moves in a straight line
Correct Answer: A
Explanation:
Work done by such force is always zero when a force of constant magnitude always at right angle to the velocity of a particle when the motion of the particle takes place in a plane.
$\therefore$ From work-energy theorem, $ \Delta K = 0$
A force $\overrightarrow F = \left( {5\overrightarrow i + 3\overrightarrow j + 2\overrightarrow k } \right)N$ is applied over a particle which displaces it from its origin to the point $\overrightarrow r = \left( {2\overrightarrow i - \overrightarrow j } \right)m.$ The work done on the particle in joules is
A.
$+10$
B.
$+7$
C.
$-7$
D.
$+13$
Correct Answer: B
Explanation:
The work done by a force on a particle is given by the dot product of the force and the displacement vector of the particle:
$W = \overrightarrow F \cdot \overrightarrow r$
We can substitute the given vectors into this expression:
$W = \overrightarrow F .\overrightarrow r $
$= \left( {5\widehat i + 3\widehat j + 2\widehat k} \right).\left( {2\widehat i - \widehat j} \right)$
A body of mass $' m ',$ acceleration uniformly from rest to $'{v_1}'$ in time ${T}$. The instantaneous power delivered to the body as a function of time is given by