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JEE • MATHEMATICS • RATE MEASURE

Rate measure practice · Easy

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 61-72 of 72 questions
61 Easy
61 of 72
Application +4 | -1
If the volume of a spherical ball is increasing at the rate of 4πcc/sec4 \pi \mathrm{cc} / \mathrm{sec}, then the rate of increase of its radius (in cm/sec), when the volume is 288πcc288 \pi \mathrm{cc},
View answer and explanation
Correct option: C
136\frac{1}{36}
Option C (136\frac{{1}}{{36}}) is correct because it is the rate of increase of the radius when the volume is 288π288 \pi cc.
62 Easy
62 of 72
Application +4 | -1
If the rate of increase of a side of a triangle inscribed in a circle of radius R\mathrm{R}, is R\mathrm{R} times the rate of increase of the measure of opposite angle, then the measure of the angle is ....... .
View answer and explanation
Correct option: C
π3\frac{\pi}{3}
Option C (π3\frac{\pi}{3}) is correct because using the sine rule and the given rate relationship, we can find the angle A.
63 Easy
63 of 72
Application +4 | -1
The rate of change of the surface area of a sphere of a sphere of radiusr,when the radius is increasing at the rate of 2 cm/sis proportional to
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Correct option: C
rr
Option C (rr) is correct because the rate of change of the surface area is directly proportional to the radius, as dSdt=16πr\frac{{dS}}{{dt}} = 16\pi r.
64 Easy
64 of 72
Application +4 | -1
A stone thrown upwards, has equation of motion s=490t−49t2.s = 490t - 49{t^2}. .Then, the maximum height reached by it ,is
View answer and explanation
Correct option: C
1225
Option C (1225) is correct because it is the maximum height reached by the stone, found by setting the velocity to zero and solving for the displacement.
65 Easy
65 of 72
Application +4 | -1
If there is an error of 2%2\% in measuring the length of a simple pendulum, then percentage error in its period is
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Correct option: A
1 %1\,\%
Option A (1 %1\,\%) is correct because the percentage error in the time period is half the percentage error in the length, as the period is proportional to the square root of the length.
66 Easy
66 of 72
Application +4 | -1
A rocket, weighing 9000 N fixed vertically upward. The gasses are expelled at a rate of 30  kg s−130\,\,{\text{kg}}\,{{\text{s}}^{ - 1}} with a speed of 0.6 km/s if the rocket weight 5000 N at the burnout, then the net upward force on the rocket at fire off should be
View answer and explanation
Correct option: B
1.8×104  N1.8 \times {10^4}\,\,N
Option B (1.8×104 N1.8 \times {10^4}\,N) is correct because the net upward force is calculated by multiplying the exhaust velocity by the rate of mass expulsion.
67 Easy
67 of 72
Application +4 | -1
The displacement of a body along the X –axis depends on time as x=t+2\sqrt x = t + 2 then the velocity of body
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Correct option: A
Increase with time
Option A (Increase with time) is correct. The velocity is directly proportional to (t+2), so it increases as time increases.
68 Easy
68 of 72
Application +4 | -1
If a particle moves according to the law s=6t2−t32s = 6{t^2} - \frac{{{t^3}}}{2}, then the time at which it is momentarily at rest is
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Correct option: C
t=0,  8t = 0,\;8
Option C (t=0,  8t = 0,\;8) is correct because the particle is momentarily at rest when its velocity (first derivative of position) is zero.
69 Easy
69 of 72
Application +4 | -1
If a particle is moving with velocity v(t)=cos⁡  πtv\left( t \right) = \cos \,\,\pi t along a straight line such that at t=0,   s=4t = 0,\;\,s = 4 its position function is given by
View answer and explanation
Correct option: C
1π  sin⁡  πt  +  4\frac{1}{\pi }\,\,\sin \,\,\pi t\,\, + \,\,4
Option C (1π sin⁡ πt + 4\frac{1}{\pi }\,\sin \,\pi t\, + \,4) is correct because it represents the position function obtained by integrating the velocity function and applying the initial condition.
70 Easy
70 of 72
Application +4 | -1
A point is moving on y=4−2x2y = 4 - 2{x^2}. The xx-coordinate of the point is decreasing at the rate of 5 unit per second. Then, the rate at which yy-coordinate of the point is changing when the point is at (1,2)is
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Correct option: D
20 units
Option D (20 units) is correct because it is calculated by differentiating the equation with respect to time and substituting the given values.
71 Easy
71 of 72
Application +4 | -1
The distance travelled by a motor car in tt seconds after the brakes are applied is s feet, where s=22t−12t2s=22 t-12 t^2. The distance travelled by the car before it stops, is
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Correct option: A
10.08  ft10.08 \mathrm\,\,{ft}
Option C (160πcm2/sec160 \pi \mathrm{cm}^2 / \mathrm{sec}) is correct because it is the rate of change of its surface area when the radius is 200 cm.
72 Easy
72 of 72
Application +4 | -1
If T=2πlgT=2 \pi \sqrt{\frac{l}{g}}, then relative errors in TT and ll are in the ratio
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Correct option: A
1/21 / 2
Option A (1/2) is correct because the relative error in T is half the relative error in l.
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