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Quiz 5
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Q.1
A particle of mass is executing oscillations about the origin on the x-axis. Its potential energy is V(x)=k|x|3, where k is a positive constant. If the amplitude of oscillation is a, then its time period T is [ AIIMS 2008]
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a)proportional to 1/√2
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b) proportional to √a
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c)independent a3/2
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d)none of these
Explanation
V(x)=k|x|3 since F=-dV(x) / dx=-3k|x|2 comparing above equation with F-mω2x we get -3k|x|2=-mω2x Thus ω2=3kx/m here x is displacement given by x=asinωtAnswer: (a)
Q.2
The average speed of the bob of a simple pendulum oscillating with a small amplitude A and time period T as... [ AIIMS 2009]
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a) 4A/T
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b) 2πA/T
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c) 4πA/T
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d)2A/T
Explanation
y=Asin(2πt /T)⇒ distance travel in time t=T/4=A ⇒ average speed=A/ (T/4)=4A/T Answer: (a)
Q.3
A pendulum is swinging in an elevator. Its periodic will be greatest when the elevator is .. [ AIIMS 20010]
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a) moving upwards at constant speed
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b) moving downwards
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c)moving downwards at constant speed
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d)accelerating downwards
Explanation
Periodic time of simple pendulum T ∝ (l/√g When the elevator is accelerating downwards, the net gravitational acceleration is ( g-a), so, the time period when elevation is accelerating downwards is greatest. Answer:(d)
Q.4
A tuning fork when sound together with a tuning fork of frequency 256 emits two two beats. On loading the tuning fork of frequency 256, the number of beats heard is 1 per second. The frequency of first tuning fork is
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a) 257
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b) 258
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c) 254
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d) 256
Explanation
The possible frequencies of unknown fork are (256±2) that is 254 and 258. On loading the fork of frequency 256, its frequency decreases. this is possible only with fork of frequency 254 Answer: (c)
Q.5
If A is the area of cross-section of spring is its length E is Young's modulus of the material of the spring then time period and force constant of the spring will be respectively .. [ AIIMS 2010]
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a)
0%
b)
0%
c)
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d)
Explanation
According to the formula for Young's Modulus E=FL/AΔL Here ΔL is increase in length of spring F=EAΔL / L Now, according to Hook's law F=kΔL where k is the spring constant From above equations kΔL=EAΔL / L k=EA/L Time period T=2π√(M/k) T=2π √(ML/EA) Answer: (d)
Q.6
The length of a spring is l and its force constant is k. When a weight W is suspended from it. its length increases by x. If the spring is cut into two equal parts and put in parallel and same weight W is suspended from then, then the extension will be ..[ MPPMT 1994]
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a)2x
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b) x
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c)x/2
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d)x/4
Explanation
Before cutting force constant is k by making two parts each have force constant 2kby connecting parallel force constant of combination k'=4k as force is same k'x'=kx 4kx'=kx x'=x/4Answer: (d)
Q.7
k is the force constant of spring. The work done in increasing its extension from l1 to l2 will be .. [ MPPMT 1994]
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a) k(l2 - l1)
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b) (k/2)(l2 - l1)
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c)k( l22 - l12 )
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d)(k/2)( l22 - l12 )
Explanation
From the formula Potential energy=½ (k) y 2 Now Work done=Change in potential energy Work done=½(k) l2 2 - ½(k) l1 2Answer: (d)
Q.8
Identify the exact statement among the following ..[ UGET 1995]
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a) The greater the mass of a pendulum bob, the shorter is its frequency of oscillation
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b) A simple pendulum with a bob of mass M swings with an angular amplitude is 20° the tension in the string is less than Mgcosθ
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c)As the length of the simple pendulum is increased, the maximum velocity of its bob during its oscillation will also increase
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d)The fraction change in the time period of a pendulum on changing the temperature is independent of the length of the pendulum
Explanation
a) Time period is independent of mass hence option is wrongb) Tension in the sting at θ=20° is T=Mgcos20 + Mu2 / L ∴ T > Mgcos20. Option is wrongc) MAx.velocity=aω=a √(g/l), as l increases ω decreases , so option is wrongd) We know that T ∝√l and l'=l(1+αΔt) Thus T'-T ∝ √(αΔt) Answer:(d)
Q.9
A particle is vibrating in simple harmonic motion with an amplitude of 4cm. At what displacement from the equilibrium position is its energy half and kinetic ? [ MNR 1995]
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a) 1cm
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b) √2 cm
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c) 2cm
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d) 2√2 cm
Explanation
Total energy=½kA2 Potential energy=½ ky2 (½kA2)=2 (½ ky2) y2=2A2 y=√(2A) y=√(2×4)=2√2Answer: (d)
Q.10
Which of the following four statements is False. [ UGET 1995]
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a) A body can have zero velocity and still be accelerated
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b) A body can have a constant velocity and still have a varying speed
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c)A body can have a constant speed and still have a varying velocity
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d)The direction of the velocity of a body can change when its acceleration is constant
Explanation
Answer: (b)
Q.11
A large horizontal surface moves up and down in simple harmonic motion with an amplitude of 1cm. If a mass of 10kg( which is placed on this surface) is to remain continuously in contact with it, the frequency of SHM should not exceed.. [ SCRA 1995]
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a) 0.5Hz
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b) 1.5Hz
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c)5.0Hz
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d)10.0Hz
Explanation
Force down ward=Force upwardmg-N=mω2AWhen N=0 object will be thrown out ∴mg=mω2Ag=ω2Aω2=g/A But ω=2πν (2πν)2=g/A Answer: (c)
Q.12
Which of the following is simple harmonic motion [ CBSE 1994]
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a) Wave moving through a string fixed at both ends
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b) earth spinning about own axis
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c)ball bouncing between two rigid vertical walls
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d)particle moving in a circle with uniform speed
Explanation
Answer:(a)
Q.13
For a particle executing simple harmonic motion, which of the following statement is not correct? [ MPPMT 1997]
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a) The total energy of the particle always remain same
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b) The restoring force is always directed towards a fixed point
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c) The restoring force is maximum at the extreme positions
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d) The acceleration of the particle is maximum at the equilibrium position
Explanation
Answer: (d)
Q.14
for a particle executing simple harmonic motion, which of the following statements is not correct? [ MPPMT 1997]
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a)The total energy of the particle always remains the same
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b) The restoring force is always directed towards fixed point
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c)The restoring force is maximum at the extreme positions
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d)The acceleration of the particle is maximum at the equilibrium position
Explanation
Answer: (d)
Q.15
The bob of a simple pendulum is displaced from its equilibrium position O to a position Q which is at height h above Q and the bob is then released. Assuming the mass of the bob to be m and time period of oscillations to be 2.0sec, the tension in the string when bob passes through O is ..[ AMU 1995]
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a)
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b)
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c)
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d)
Explanation
Period of oscillation T=2 secTension in the string at lowest point is F=mg + mv2 / l --eq(1)According to law of conservation of energy ½ mv2=mgh v2=2gh Thus substituting value of l and v2 in equation 1 we get Answer: (a)
Q.16
A uniform circular disc of mass 12kg is held by two identical springs as shown in figure. When the disc is slightly pressed down and released, it executes S.H.M of period 2s. The force constant of each spring is [ AMU 1998]
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a)236 Nm-1
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b) 118.3 Nm-1
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c)59.15 Nm-1
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d)none of these
Explanation
T=2π√(M/k) 4=4π2 (12/k) k=12 π2 k=118.31 this restoring force constant is of two spring for one spring restoring force constant=59.15 Nm-1 Answer:(c)
Q.17
A pendulum clock is set to give correct time at the sea level. this clock is moved to hill station at an altitude of 2500 m above the sea level. In order to keep correct time on the hill station, the length of the pendulum [ UPSE 1994]
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a) has to be reduced
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b) has to be increased
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c) need no adjustment
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d) need no adjustment but its mass has to be increased
Explanation
At altitude value of g will reduce thus length also should be reduced Answer: (a)
Q.18
A particle moving along X-axis, executes simple harmonic motion. Then the force acting on it is given by [ CBSE 1994] where A and k are positive constants
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a)-Akx
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b) Acos(kx)
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c)Aexp(-kx)
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d)Akx
Explanation
Answer: (a)
Q.19
A system exhibiting SHM must posses ..[ CET 1994]
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a) inertia only
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b) elasticity as well as inertia
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c)elasticity, inertia and an external force
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d)elasticity only
Explanation
Answer: (c)
Q.20
Two spring constant 1500N/m and 3000N/m respectively are stretched with the same force. They will have potential energy in the ratio.[ MPPMT 1998]
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a) 4:1
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b) 1:4
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c)2:1
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d)1:2
Explanation
potential energy U=½ ky2 Now F=kyy=F/kU=F2 /2kU1 / U2=k2/k1U1 / U2=3000 / 1500=2:1 Answer:(c)
Q.21
The displacement of a particle from its mean position (in m) varies with time according to the relation y=0.2sin(10πt + 1.5π)cos(10πt + 1.5π)The motion of the particle is .. [ CPMT 1998]
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a) Not simple harmonic
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b) Simple harmonic with time period 0.2 sec
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c) Simple harmonic with time period 0.1 sec
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d) Along a circular path
Explanation
y=0.2sin(10πt + 1.5π)cos(10πt + 1.5π) y=0.1[2sin(10πt + 1.5π)cos(10πt + 1.5π)] y=0.1 sin2(10πt + 1.5π) y=0.1sin(20πt + 3π) Comparing above equation with y=0.1sin(ωt+Φ) ω=20π or 2π/T=20π T=1/10=0.1 Option 'c' is correct Answer: (c)
Q.22
Three springs each of force constant k are connected at equal angles with respect to each other to a common mass M. The other end of the springs is rigidly fixed. If the mass is pulled towards any one of the springs, then the period of oscillation will be
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a)
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b)
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c)
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d)
Explanation
Suppose the mass M is pushed down in the direction AO by 'y' then the spring S1 extends by y while spring S2 and S3 are compressed by ycos60 and ycos60 ( because the angle between any two spring is 120°). ThEn the total restoring force on mass M along AO is F=ky + (ky/2)cos60 + (ky/2)cos60 F=ky + (ky/4) + (ky/4)=3ky/2Hence, equivalent spring constant=3k/2 T=2π√(2M/3k)Answer: (b)
Q.23
A load of mass m falls from a height h onto the scale pan hung from a spring as shown in the adjoining figure. If the spring constant is k and mass of the scale pan is zero and the mass m does not bounce relative to pan, then the amplitude of the vibration is
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a) mg/k
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b)
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c)
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d)
Explanation
Let the extension in the spring be x. Then loss in P.E of mass m=gain in K.E of spring mg(h+x)=½ kx2 or kx2 -2mgx -2mgh=0 roots are Answer: (b)
Q.24
A tuning fork of frequency 380Hz is moving towards a wall with a velocity of 4m/s. Calculate the number of beats heard per second between the direct and the reflected sounds. Velocity of sound in air is 340 m/sec ..[CMET 1995]
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a)0
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b)7
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c)5
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d)10
Explanation
As indicated in question the observer is between the source and the wall. Both the direct and reflected waves have same frequency.No beats will be heard Answer:(a)
Q.25
A source X of unknown frequency produces 8 beats with a source of 250Hz and 12 beats with a source of 270Hz and 12 beats with a source of 270 Hz. The frequency of source X is
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a)258 Hz
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b) 242Hz
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c)262Hz
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d)282Hz
Explanation
of unknown frequency=250±8 that is 242 or 258 It produces 12 beats with tuning fork of frequency 270. Therefore the correct frequency of X is 258 Answer: (a)
Q.26
in the production of beats by the two waves of same amplitudes and nearly same frequencies, the maximum loudness heard corresponding to each of the constituent wave is
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a) same
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b) 2 times
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c)4 times
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d)8 times
Explanation
I max=(a1+a2)2 I max=4a2=4IAnswer: (c)
Q.27
When a tuning fork A of unknown frequency is sounded with another tuning fork B of frequency 256Hz, then 3 beats per second are observed. After that A is loaded with wax and sounded, then again 3 beats per second are observed. The frequency of the tuning fork A is .. [ MPPMT 1994]
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a) 250 Hz
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b) 253 Hz
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c)259Hz
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d)262Hz
Explanation
frequency of A=256±3=259 or 253.On loading the frequency of A decreases. Since again 3 beats are heard, the frequency of a is 259 which reduces to 253 on loading. Answer:(b)
Q.28
Two waves having intensities in the ratio of 9:1 produce interference. The ratio of maximum to minimum intensity is equal to : [MNR 1987]
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a) 10:8
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b) 9:1
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c) 4:1
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d) 2:1
Explanation
Answer: (c)
Q.29
A number of tuning forks are arranged in the order of increasing frequency and any two successive tuning fork produce 4 beats per second, when sounded together. If the last tuning fork has frequency octave higher than that of the first tuning fork and the frequency of the first tuning fork is 256 hz, then the number of tuning forks is [ CPMT 1998]
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a)63
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b) 64
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c)65
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d)66
Explanation
Let the frequency of first tuning fork be 'n', then frequency of last tuning fork is 2nThe frequencies are thus n, n+4, n+8,...,2n.Now L=A+(N-1)d Here A is frequency of first tuning fork, N=Number of tuning forkd=successive difference in frequency∴ 2n=n +( N-1)4 on solvingN=65Answer: (c)
Q.30
When two tuning forks of nearly same frequency are sounded to gather they produce beats, the velocity of the propagation of beats
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a) Is greater than the velocity of sound
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b) Is smaller than the velocity of sound
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c)Depends on the relative frequencies
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d)same as the speed of sound
Explanation
Answer: (d)
0 h : 0 m : 1 s
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