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JEE • MATHEMATICS • RATE MEASURE

Rate measure practice · Mixed difficulty

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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 91-110 of 110 questions
91 Easy
91 of 110
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.
92 Easy
92 of 110
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.
93 Medium
93 of 110
Application +4 | -1
For a particle moving in a straight line, if time tt be regarded as a function of velocity vv, then the rate of change of the acceleration a is given by
View answer and explanation
Correct option: C
−a3d2tdv2- {a^3}\frac{{{d^2}t}}{{d{v^2}}}
Option C (−a3d2tdv2- {a^3}\frac{{{d^2}t}}{{d{v^2}}}) is correct because it represents the rate of change of acceleration with respect to time, derived by treating time as a function of velocity.
94 Easy
94 of 110
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
View answer and explanation
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.
95 Medium
95 of 110
Application +4 | -1
The radius of a circular plate is increasing at the rate of 0.01 cm/s when the radius is 12 cm. Then , The rate at which the area increase, is
View answer and explanation
Correct option: A
0.24  π  0.24\;\pi \; sq cm/s
Option A (0.24  π  sqcm/s)iscorrectbecauseitrepresentstherateofchangeoftheareacalculatedusingthegivenrateofchangeoftheradius.0.24\;\pi \; sq cm/s) is correct because it represents the rate of change of the area calculated using the given rate of change of the radius.
96 Medium
96 of 110
Application +4 | -1
A point on the parabola y2=18x{y^2} = 18x at which the ordinate increases at twice the rate of the abscissa, is
View answer and explanation
Correct option: D
(98,9  2)  \left( {\frac{9}{8},\frac{9}{{\;2}}} \right)\;
Option D (98,9  2)  \left( {\frac{9}{8},\frac{9}{{\;2}}} \right)\; is correct because it satisfies the given condition where the ordinate increases at twice the rate of the abscissa on the parabola.
97 Easy
97 of 110
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.
98 Easy
98 of 110
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.
99 Easy
99 of 110
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.
100 Easy
100 of 110
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
View answer and explanation
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.
101 Easy
101 of 110
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
View answer and explanation
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.
102 Medium
102 of 110
Application +4 | -1
If the distance s covered by a particle in time t is proportional to the cube root of its velocity, then the acceleration is
View answer and explanation
Correct option: D
∝s5\propto {s^5}
Option D (∝s5\propto {s^5}) is correct because the acceleration is proportional to the fifth power of the distance covered, derived from the given relationship between distance and velocity.
103 Easy
103 of 110
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.
104 Hard
104 of 110
Application +4 | -1
A particle moves on the parabola y2=4ax{y^2} = 4ax in such a way that its projection on the yy-axis has a constant velocity. Then its projection on xx-axis moves with
View answer and explanation
Correct option: B
Constant acceleration
Option B (Constant acceleration) is correct because the second derivative of the x-coordinate with respect to time is constant, given the constant velocity of the y-projection.
105 Medium
105 of 110
Application +4 | -1
If y=x+c1+x2y = \frac{{x + c}}{{1 + {x^2}}} where cc is a constant, then when yy is stationary, xyxy is equal to
View answer and explanation
Correct option: A
12\frac{1}{2}
The correct answer is 12\frac{1}{2} because it is derived by finding the stationary points of the function and then calculating the value of xyxy at those points.
106 Easy
106 of 110
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
View answer and explanation
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.
107 Easy
107 of 110
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
View answer and explanation
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.
108 Easy
108 of 110
Application +4 | -1
If T=2πlgT=2 \pi \sqrt{\frac{l}{g}}, then relative errors in TT and ll are in the ratio
View answer and explanation
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.
109 Medium
109 of 110
Application +4 | -1
If the velocity vv of a particle moving along a straight line and its distance ss from a fixed point on the line are related by υ2=a2+s2,{\upsilon ^2} = {a^2} + {s^2}, then its acceleration equals
View answer and explanation
Correct option: B
s
Option B (s) is correct because the acceleration is equal to ss when υ2=a2+s2{\upsilon ^2} = {a^2} + {s^2}.
110 Medium
110 of 110
Application +4 | -1
The radius and height of a cylinder are equal. If the radius of the sphere is equal to the height of the cylinder, then the ratio of the rates of increase of the volume of the sphere and the volume of the cylinder is
View answer and explanation
Correct option: A
4:34:3
Option A (4:3) is correct because it represents the ratio of the rates of increase of the volume of the sphere and the volume of the cylinder when their dimensions are related as described.
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