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JEE Advanced · Physics · Laws of Motion

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1
A person sitting inside an elevator performs a weighing experiment with an object of mass 50 kg. Suppose that the variation of the height yy (in m) of the elevator, from the ground, with time tt (in s) is given by y=8[1+sin⁡(2πtT)]y=8\left[1+\sin\left(\dfrac{2\pi t}{T}\right)\right], where T=40πT=40\pi s. Taking acceleration due to gravity, g=10 m/s2g=10\text{ m/s}^2, the maximum variation of the object's weight (in N) as observed in the experiment is ____
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Your answer: —
Correct answer: 2
2
A projectile of mass 200 g is launched in a viscous medium at an angle 60∘60^\circ with the horizontal, with an initial velocity of 270 m/s. It experiences a viscous drag force F⃗=−cv⃗\vec F=-c\vec v where the drag coefficient c=0.1c=0.1 kg/s and v⃗\vec v is the instantaneous velocity of the projectile. The projectile hits a vertical wall after 2 s. Taking e=2.7e=2.7, the horizontal distance of the wall from the point of projection (in m) is ____
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Your answer: —
Correct answer: 170
3
A particle of mass mm is moving in the xyxy-plane such that its velocity at a point (x,y)(x, y) is given as v⃗=α(yx^+2xy^)\vec v = \alpha(y\hat x + 2x\hat y), where α\alpha is a non-zero constant. What is the force F⃗\vec F acting on the particle?
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A
F⃗=2mα2(xx^+yy^)\vec F = 2m\alpha^2(x\hat x + y\hat y)
B
F⃗=mα2(yx^+2xy^)\vec F = m\alpha^2(y\hat x + 2x\hat y)
C
F⃗=2mα2(yx^+xy^)\vec F = 2m\alpha^2(y\hat x + x\hat y)
D
F⃗=mα2(xx^+2yy^)\vec F = m\alpha^2(x\hat x + 2y\hat y)
4
A student skates up a ramp that makes an angle 30∘30^\circ with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed v0v_0 and wants to turn around over a semicircular path xyzxyz of radius RR, during which he/she reaches a maximum height hh (at point yy) from the ground, as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then (gg is the acceleration due to gravity)
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A
v02−2gh=12gRv_0^2-2gh=\dfrac12gR
B
v02−2gh=32gRv_0^2-2gh=\dfrac{\sqrt3}{2}gR
C
the centripetal force required at points xx and zz is zero
D
the centripetal force required is maximum at points xx and zz
5
Put a uniform meter scale horizontally on your extended index fingers, with the left one at 0.000.00 cm and the right one at 90.0090.00 cm. When you attempt to move both the fingers slowly towards the center, initially only the left finger slips with respect to the scale and the right finger does not. After some distance, the left finger stops and the right one starts slipping. Then the right finger stops at a distance xRx_R from the center (50.0050.00 cm) of the scale, and the left one starts slipping again. This happens because of the difference in the frictional forces on the two fingers. If the coefficients of static and dynamic friction between the fingers and the scale are 0.400.40 and 0.320.32 respectively, the value of xRx_R (in cm) is ______.
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Your answer: —
Correct answer: 25.6
6
A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m=0.4 kgm=0.4\ kg is at rest on this surface. An impulse of 1.0 Ns1.0\ Ns is applied to the block at time t=0t=0 so that it starts moving along the xx-axis with a velocity v(t)=v0e−t/τv(t) = v_0 e^{-t/\tau}, where v0v_0 is a constant and τ=4 s\tau = 4\ s. The displacement of the block, in metersmeters, at t=τt=\tau is __________. Take e−1=0.37e^{-1}=0.37.
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Your answer: —
Correct answer: 6.3
7
Paragraph\textbf{Paragraph}: A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity ω\omega is an example of a non-inertial frame of reference. The relationship between the force F⃗rot\vec F_{\text{rot}} experienced by a particle of mass mm moving on the rotating disc and the force F⃗in\vec F_{\text{in}} experienced by the particle in an inertial frame of reference is F⃗rot=F⃗in+2m(v⃗rot×ω⃗)+m(ω⃗×r⃗)×ω⃗\vec F_{\text{rot}}=\vec F_{\text{in}}+2m(\vec v_{\text{rot}}\times\vec\omega)+m(\vec\omega\times\vec r)\times\vec\omega, where v⃗rot\vec v_{\text{rot}} is the velocity of the particle in the rotating frame of reference and r⃗\vec r is the position vector of the particle with respect to the centre of the disc. Now consider a smooth slot along a diameter of a disc of radius RR rotating counter-clockwise with a constant angular speed ω\omega about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the xx-axis along the slot, the yy-axis perpendicular to the slot and the zz-axis along the rotation axis (ω⃗=ωk^\vec\omega=\omega\hat k). A small block of mass mm is gently placed in the slot at r⃗=(R/2)i^\vec r=(R/2)\hat i at t=0t=0 and is constrained to move only along the slot. The distance rr of the block at time tt is
Not attempted
A
R4(eωt+e−ωt)\dfrac{R}{4}\left(e^{\omega t}+e^{-\omega t}\right)
B
R2cos⁡ωt\dfrac{R}{2}\cos\omega t
C
R4(e2ωt+e−2ωt)\dfrac{R}{4}\left(e^{2\omega t}+e^{-2\omega t}\right)
D
R2cos⁡2ωt\dfrac{R}{2}\cos2\omega t
8
Paragraph\textbf{Paragraph}: A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity ω\omega is an example of a non-inertial frame of reference. The relationship between the force F⃗rot\vec F_{\text{rot}} experienced by a particle of mass mm moving on the rotating disc and the force F⃗in\vec F_{\text{in}} experienced by the particle in an inertial frame of reference is F⃗rot=F⃗in+2m(v⃗rot×ω⃗)+m(ω⃗×r⃗)×ω⃗\vec F_{\text{rot}}=\vec F_{\text{in}}+2m(\vec v_{\text{rot}}\times\vec\omega)+m(\vec\omega\times\vec r)\times\vec\omega, where v⃗rot\vec v_{\text{rot}} is the velocity of the particle in the rotating frame of reference and r⃗\vec r is the position vector of the particle with respect to the centre of the disc. Now consider a smooth slot along a diameter of a disc of radius RR rotating counter-clockwise with a constant angular speed ω\omega about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the xx-axis along the slot, the yy-axis perpendicular to the slot and the zz-axis along the rotation axis (ω⃗=ωk^\vec\omega=\omega\hat k). A small block of mass mm is gently placed in the slot at r⃗=(R/2)i^\vec r=(R/2)\hat i at t=0t=0 and is constrained to move only along the slot. The net reaction of the disc on the block is
Not attempted
A
12mω2R(e2ωt−e−2ωt)j^+mgk^\dfrac12m\omega^2R\left(e^{2\omega t}-e^{-2\omega t}\right)\hat j+mg\hat k
B
12mω2R(eωt−e−ωt)j^+mgk^\dfrac12m\omega^2R\left(e^{\omega t}-e^{-\omega t}\right)\hat j+mg\hat k
C
−mω2Rcos⁡ωt j^−mgk^-m\omega^2R\cos\omega t\,\hat j-mg\hat k
D
mω2Rsin⁡ωt j^−mgk^m\omega^2R\sin\omega t\,\hat j-mg\hat k
9
A wire, which passes through the hole in a small bead, is bent in the form of a quarter of a circle. The wire is fixed vertically on the ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from AA to BB, the force it applies on the wire is
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A
always radially outwards.
B
always radially inwards.
C
radially outwards initially and radially inwards later.
D
radially inwards initially and radially outwards later.
10
A block of mass m1=1 kgm_1=1\text{ kg} and another mass m2=2 kgm_2=2\text{ kg} are placed together (see figure) on an inclined plane with angle of inclination θ\theta. Various values of θ\theta are given in List-I. The coefficient of friction between block m1m_1 and the plane is always zero. The coefficient of static and dynamic (kinetic) friction between block m2m_2 and the plane are equal, μ=0.3\mu=0.3. List-II gives expressions for the friction force on block m2m_2. Match the correct expression in List-II with the angles given in List-I. The acceleration due to gravity is gg. [Useful information: tan⁡(5.5°)≈0.1\tan(5.5°)\approx0.1; tan⁡(11.5°)≈0.2\tan(11.5°)\approx0.2; tan⁡(16.5°)≈0.3\tan(16.5°)\approx0.3] List I\textbf{List I} P.\textbf{P.} θ=5°\theta=5° Q.\textbf{Q.} θ=10°\theta=10° R.\textbf{R.} θ=15°\theta=15° S.\textbf{S.} θ=20°\theta=20° List II\textbf{List II} 1.\textbf{1.} m2gsin⁡θm_2g\sin\theta 2.\textbf{2.} (m1+m2)gsin⁡θ(m_1+m_2)g\sin\theta 3.\textbf{3.} μm2gcos⁡θ\mu m_2g\cos\theta 4.\textbf{4.} μ(m1+m2)gcos⁡θ\mu(m_1+m_2)g\cos\theta
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A
P-1, Q-1, R-1, S-3
B
P-2, Q-2, R-2, S-3
C
P-2, Q-2, R-2, S-4
D
P-2, Q-2, R-3, S-3
11
A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance of 1.2 m1.2\text{ m} from the person. In the following, the state of the lift's motion is given in List-I and the distance where the water jet hits the floor of the lift is given in List-II. Match the statements in List-I with those in List-II. List I\textbf{List I} P.\textbf{P.} Lift is accelerating vertically up. Q.\textbf{Q.} Lift is accelerating vertically down with acceleration less than the gravitational acceleration. R.\textbf{R.} Lift is moving vertically up with constant speed. S.\textbf{S.} Lift is falling freely. List II\textbf{List II} 1.\textbf{1.} d=1.2 md=1.2\text{ m} 2.\textbf{2.} d>1.2 md>1.2\text{ m} 3.\textbf{3.} d<1.2 md<1.2\text{ m} 4.\textbf{4.} No water leaks out of the jar
Not attempted
A
P-2, Q-3, R-2, S-4
B
P-2, Q-3, R-1, S-4
C
P-1, Q-1, R-1, S-4
D
P-2, Q-3, R-1, S-1
12
A rocket is moving in a gravity-free space with a constant acceleration of 2 m s−22\ \text{m s}^{-2} along the +x+x direction (see figure). The length of a chamber inside the rocket is 4 m4\ \text{m}. A ball is thrown from the left end of the chamber in the +x+x direction with a speed of 0.3 m s−10.3\ \text{m s}^{-1} relative to the rocket. At the same time, another ball is thrown in the −x-x direction with a speed of 0.2 m s−10.2\ \text{m s}^{-1} from the right end, relative to the rocket. The time in seconds when the two balls hit each other is
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Your answer: —
Correct answer: 8
13
A bob of mass mm, suspended by a string of length l1l_1, is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass mm suspended by a string of length l2l_2, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision, acquires the minimum speed required to complete a full circle in the vertical plane, the ratio l1/l2l_1/l_2 is
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Your answer: —
Correct answer: 5
Minimum speeds for a vertical circle: vbottom=5glv_{bottom}=\sqrt{5gl}, vtop=glv_{top}=\sqrt{gl}. An elastic collision between equal masses (one at rest) exchanges velocities.
Deconstruction\textbf{Deconstruction}: Bob 1 (string length l1l_1) is launched with the minimum speed for a full vertical circle. At its highest point it elastically collides (equal masses) with bob 2 (string length l2l_2, initially at rest). After the collision bob 2 also has exactly the minimum speed needed for its own full vertical circle. Find l1/l2l_1/l_2. Strategy\textbf{Strategy}: Use the standard minimum-speed-for-vertical-circle results (vbottom=5glv_{bottom}=\sqrt{5gl}, vtop=glv_{top}=\sqrt{gl}) together with the fact that an elastic collision between equal masses (one initially at rest) exchanges velocities. Execution\textbf{Execution}: Bob 1's minimum speed at the bottom is 5gl1\sqrt{5gl_1}; by energy conservation its speed at the top of its circle is v1=gl1v_1=\sqrt{gl_1}. Since the collision is elastic and both bobs have equal mass mm with bob 2 initially at rest, the bobs exchange velocities: bob 2 receives speed v1=gl1v_1=\sqrt{gl_1} (bob 1 stops). This transferred speed is bob 2's speed at the bottom of its own circle (bob 2 hangs at rest before the collision). For bob 2 to just complete a full vertical circle, this bottom speed must equal 5gl2\sqrt{5gl_2}. So gl1=5gl2\sqrt{gl_1}=\sqrt{5gl_2}, giving l1=5l2l_1=5l_2, i.e. l1/l2=5l_1/l_2=5. Verification\textbf{Verification}: Dimensionally l1/l2l_1/l_2 is a pure number, consistent with a ratio of lengths. Cross-checked independently against the officially published FIITJEE solution to this exact problem, which arrives at the same value via the identical method (note: one third-party web-search summary encountered during verification claimed 1/51/5, but its own shown algebra contained an internal arithmetic inconsistency — it wrote \"5gl1=5gl25gl_1=5gl_2\" yet concluded a ratio of 1/51/5 instead of 11; that source's arithmetic is not trusted, and the from-scratch derivation together with the primary FIITJEE solution and this paper's own printed answer key all agree on 55). Final Answer\textbf{Final Answer}: l1/l2=5l_1/l_2 = 5.