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JEE • MATHEMATICS • RATE MEASURE

Rate measure practice · Mixed difficulty

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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 31-60 of 110 questions

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31 Easy
31 of 110
Application +4 | -1
The rate of change of surface area of a sphere of raditis rr when the radius is increasing at the rate of 2 cm/sec2 \mathrm{~cm} / \mathrm{sec} is proportional to
View answer and explanation
Correct option: D
rr
The correct answer is D (rr) because the rate of change of the surface area of a sphere is directly proportional to its radius when the radius is increasing at a constant rate.
32 Easy
32 of 110
Application +4 | -1
Rays from the sun are focused by a lens of diameter 5m on to a block of ice (l=80 kcal/10g)\left( {l = 80\,kcal/10g} \right) and 10 g of ice is melted in 20 minutes. Calculate the heat received from the sun on 1 cm21\,c{m^2} per minute
View answer and explanation
Correct option: D
2 cal
Option D (2 cal) is correct because it is the heat received per unit area per minute, calculated from the latent heat of fusion and the area of the lens.
33 Easy
33 of 110
Application +4 | -1
A particle moves along the curve y=x2+2xy = {x^2} + 2x. Then, The point on the curve such that x and yx\,and\,y coordinates of the particle change with the same rate is
View answer and explanation
Correct option: C
(−12,−34)\left( { - \frac{1}{2}, - \frac{3}{4}} \right)
Option C (−12,−34)\left( { - \frac{{1}}{{2}}, - \frac{{3}}{{4}}} \right) is correct because it satisfies the condition where the rates of change of x and y coordinates are equal.
34 Easy
34 of 110
Application +4 | -1
A vehicle of 200 kg moving with a velocity of 6 m/sec . To stop if in 115sec⁡\frac{1}{{15}}\sec the required force in opposite direction
View answer and explanation
Correct option: D
8000 N
Option D (8000 N) is correct because the required force is calculated using the change in momentum divided by the time interval.
35 Medium
35 of 110
Application +4 | -1
The rate of change of area of a square SS is equal to its side length. The rate of change of side of SS is same as that of a cube CC. The rate of change of volume of CC when its side length is 2 units, is
View answer and explanation
Correct option: C
6
Option C (6) is correct because it is the rate of change of the volume of the cube when its side length is 2 units, calculated using related rates.
36 Easy
36 of 110
Application +4 | -1
The distance covered by a particle in tt second is given by x=3+8t−4t2.  x = 3 + 8t - 4{t^2}.\; After 1s its velocity will be
View answer and explanation
Correct option: A
0 unit
Option A (0 unit) is correct because the velocity after 1 second is calculated by differentiating the position function and substituting t=1.
37 Medium
37 of 110
Application +4 | -1
A balloon is pumped at the rate of cm3/\mathrm{cm}^3 / minute. The rate of increase of its surface area when the radius is b cm\mathrm{cm}, is -
View answer and explanation
Correct option: C
(2a/b)cm2/min(2 \mathrm{a} / \mathrm{b}) \mathrm{cm}^2 / \mathrm{min}
Option C ((2a/b) cm2/min(2a / b) \text{{ cm}}^2 / \text{{min}}) is correct because differentiating the volume and surface area formulas with respect to time and then substituting r=b leads to this rate of increase.
38 Easy
38 of 110
Application +4 | -1
the speed v of particle moving along a straight line is given by a+y2=x2a + {y^2} = {x^2} (where xx is its distance from the origin).The acceleration of the particle is
View answer and explanation
Correct option: C
xb\frac{x}{b}
Option B ((−∞,  0- \infty ,\;0)) is correct because the function is increasing when its derivative is positive, which occurs for x < 0.
39 Medium
39 of 110
Application +4 | -1
A particle is moving in a straight line. At time tt, the distance between the particle from its starting point is given by x=t−6t2+t3x = t - 6{t^2} + {t^3} .Its acceleration will be zero at
View answer and explanation
Correct option: B
t=2t = 2 unit time
Option B (t=2t = 2 unit time) is correct because it's the time at which the second derivative of the position function (acceleration) is zero.
40 Medium
40 of 110
Application +4 | -1
If the rate of change of volume of a sphere is equal to the rate of change of its radius, then its radius is equal to
View answer and explanation
Correct option: D
12π\frac{1}{2 \sqrt{\pi}} unit
Option A (10.08 ft) is correct because it is the distance traveled by the car before it stops, calculated using the given equation.
41 Medium
41 of 110
Application +4 | -1
The sides of an equilateral triangle are increasing at the rate of 2cm/s.The rate at . Which the area increases, when the side is 10cm is
View answer and explanation
Correct option: C
103cm2/s10\sqrt 3 \mathrm{c}{\mathrm{m}^2}/\mathrm{s}
Option C (103cm2/s10\sqrt 3 \mathrm{c}{\mathrm{m}^2}/\mathrm{s}) is correct because it represents the rate of change of the area of the equilateral triangle when the side is 10cm, calculated using related rates.
42 Easy
42 of 110
Application +4 | -1
A cone whose height is always equal to its diameter is increasing in volume at the rate of 40 cm3/sec40 \mathrm{~cm}^3 / \mathrm{sec}. At what rate is the radius increasing when its circular base area is 1 m21 \mathrm{~m}^2 ?
View answer and explanation
Correct option: D
0.002 cm/sec0.002 \mathrm{~cm} / \mathrm{sec}
The correct answer is D (0.002 cm/sec0.002 \mathrm{~cm} / \mathrm{sec}) because it is the rate of change of the radius calculated using related rates and the given volume increase.
43 Medium
43 of 110
Application +4 | -1
A spherical balloon is being inflated at the rate of 30 cc    /minc{c^{\;\;}}/\mathrm{min}. The rate of increase of the surface area of the balloon when its diameter is 14cm, is
View answer and explanation
Correct option: B
10 sq cm/min
Option B (10 sq cm/min) is correct because it is the rate of increase of the surface area calculated using related rates, given the rate of inflation and the diameter.
44 Medium
44 of 110
Application +4 | -1
A spherical raindrop evaporates at a rate proportional to its surface area at any instant tt. The differential equation giving the rate of change of the radius of the raindrop is
View answer and explanation
Correct option: A
drdt+k=0   (k>0)\frac{d r}{d t}+k=0\,\,\,(k > 0)
Option A (drdt+k=0\frac{d r}{d t}+k=0, k>0k > 0) is correct because the rate of change of the radius is proportional to the surface area, and the radius is decreasing.
45 Medium
45 of 110
Application +4 | -1
Moving along the x-axis there are two points with x=10+6t,x=3+t2.x = 10 + 6t,x = 3 + {t^2}. The speed with which they are reaching from each other at the time of encounter is in cm and t is in seconds)
View answer and explanation
Correct option: C
8 cm/s
Option C (8 cm/s) is correct because it represents the relative speed between the two points at the time of their encounter, calculated by finding the difference between their individual velocities.
46 Medium
46 of 110
Application +4 | -1
A point moves along the curve 12y=x312y = {x^3} in such a way that the rate of increase of its ordinate is more than the rate of increase of abscissa. The abscissa of the point lies in the interval
View answer and explanation
Correct option: B
(−∞,  −2)∪(2,  ∞)\left( { - \infty ,\; - 2} \right) \cup \left( {2,\;\infty } \right)
Option B ((−∞,  −2)∪(2,  ∞)\left( { - \infty ,\; - 2} \right) \cup \left( {2,\;\infty } \right)) is correct because it represents the interval where the rate of increase of the ordinate is greater than the rate of increase of the abscissa.
47 Easy
47 of 110
Application +4 | -1
A sphere of 10 cm10 \mathrm{~cm} radius has a uniform thickness of ice around it. If ice is melting at the rate of 50 cm3/min50 \mathrm{~cm}^3 / \mathrm{min} when thickness is 5 cm5 \mathrm{~cm}, then rate of change of thickness is
View answer and explanation
Correct option: B
118π\frac{1}{18 \pi}
The correct answer is 118π\frac{1}{18 \pi} because this is the calculated rate of change of the ice thickness based on the given melting rate and thickness.
48 Easy
48 of 110
Application +4 | -1
A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3/min⁡.50\,c{m^3}/\min . When the thickness of ice is 5 cm, then the rate (in cm / min) at which of the thickness of ice decreases, is
View answer and explanation
Correct option: D
118π\frac{1}{{18\pi }}
Option D (118π\frac{1}{{18\pi }}) is correct because it is the rate at which the thickness of the ice decreases when the thickness is 5 cm, calculated using related rates.
49 Easy
49 of 110
Application +4 | -1
A particle moves in a straight line so that s=ts = \sqrt t, then its acceleration is proportional to
View answer and explanation
Correct option: A
(velocity)3{(\mathrm{velocity})^3}
Option A ((velocity)3{(\mathrm{velocity})^3}) is correct because the acceleration is proportional to the cube of the velocity, derived from the given relation s=ts = \sqrt{t}.
50 Easy
50 of 110
Application +4 | -1
If a circular iron sheet of radius 30 cm30 \mathrm{~cm} is heated such that its area increases at the uniform rate of 6πcm2/hr6 \pi \mathrm{cm}^2 / \mathrm{hr}, then the rate (in mm/hr\mathrm{mm} / \mathrm{hr} ) at which the radius of the circular sheet increases is
View answer and explanation
Correct option: B
0.1
Option B (0.1) is correct because the rate at which the radius of the circular sheet increases is calculated to be 0.1 mm/hr.
51 Easy
51 of 110
Application +4 | -1
A solid cylinder of height HH has a conical portion of same height and radius 1/3rd1 / 3 r d of height removed from it. Rain water is accumulating in it, at the rate equal to π\pi times the instantaneous radius of the water surface inside the hole, the time after which hole will filled with water is
View answer and explanation
Correct option: C
H26\frac{H^2}{6}
Option C H26\frac{H^2}{6} is correct because it is the time calculated by equating the rate of water accumulation to the volume of the conical portion.
52 Easy
52 of 110
Application +4 | -1
The equation of motion of a particle moving along a straight line is s=2t3−9t2+12t,s = 2{t^3} - 9{t^2} + 12t, where the units of ss and tt are centimeter and second. The acceleration of the particle will be zero after
View answer and explanation
Correct option: A
32s\frac{3}{2}s
Option A (32s\frac{3}{2}s) is correct because it is the time at which the second derivative of the position function (acceleration) is zero.
53 Medium
53 of 110
Application +4 | -1
A particle moves along a straight line with the law of motion given by s2=at2+2bt+c.{s^2} = a{t^2} + 2bt + c. then the acceleration varies are
View answer and explanation
Correct option: A
1s3\frac{1}{{{s^3}}}
Option A (1s3\frac{1}{{{s^3}}}) is correct because the acceleration is inversely proportional to the cube of the displacement, derived from differentiating the given equation of motion.
54 Easy
54 of 110
Application +4 | -1
A particle moves along a straight line according to the law s=16−2t+3t3s = 16 - 2t + 3{t^3} , where ss a metre is the distance of the particle from a fixed point at the end of tt seconds. The acceleration of the particle at the end of 2s2s is
View answer and explanation
Correct option: A
36m/s236m/{s^2}
Option A (36m/s236m/{s^2}) is correct because it is the acceleration calculated by taking the second derivative of the position function and evaluating it at t=2s.
55 Easy
55 of 110
Application +4 | -1
The points on the curve 12y=x312y = {x^3} whose ordinate and abscissa change at the same rate, are
View answer and explanation
Correct option: A
(−2,  −2/3),(2,  2/3)\left( { - 2,\; - 2/3} \right),\left( {2,\;2/3} \right)
Option A ((−2,  −2/3),(2,  2/3)\left( { - 2,\; - 2/3} \right),\left( {2,\;2/3} \right)) is correct because these are the points on the curve where the rate of change of the ordinate and abscissa are equal.
56 Easy
56 of 110
Application +4 | -1
The temperature of 100 g of water is to be raised from 24∘C24^\circ C to 90∘C90^\circ C by adding steam to it. Calculate the mass of the steam required for the purpose
View answer and explanation
Correct option: B
12 g
Option B (12 g) is correct because it applies the principle of calorimetry, equating the heat gained by the water to the heat lost by the steam.
57 Easy
57 of 110
Application +4 | -1
The semi-vertical angle of a cone is 45∘45^{\circ}. If the height of the cone is 20.025, then its approximate lateral surface area, is
View answer and explanation
Correct option: A
4012π401 \sqrt{2} \pi
Option A (4012π401 \sqrt{{2}} \pi) is correct because it represents the approximate lateral surface area of the cone, calculated using differentials.
58 Medium
58 of 110
Application +4 | -1
If the rate of change of area of a square plate is equal to that of the rate of change of its perimeter, then length of the side is
View answer and explanation
Correct option: B
2 units
Option D (f(6)≥8f(6) \ge 8) is correct because it is the lower bound for f(6) based on the Mean Value Theorem and the given conditions.
59 Easy
59 of 110
Application +4 | -1
A stone is dropped into a quiet lake. If the waves moves in circles at the rate of 30cm/sec when the radius is 50 m, the rate of increase of enclosed area is
View answer and explanation
Correct option: A
30  π  m2/sec30\;\pi \;{\mathrm{m}^2}/\mathrm{sec}
Option A (30  π  m2/sec30 \;\pi \;{\mathrm{m}^2}/\mathrm{sec}) is correct because the rate of increase of the enclosed area is calculated using the formula dAdt=2πrdrdt\frac{{dA}}{{dt}} = 2\pi r\frac{{dr}}{{dt}}.
60 Medium
60 of 110
Application +4 | -1
The edge of a cube is equal to the radius of a sphere. If the edge and the radius increase at the same rate, then the ratio of the increases in surface areas of the cube and sphere is
View answer and explanation
Correct option: B
3  :2π3\;:2\pi
Option B (3  :2π3\;:2\pi) is correct because it is the ratio of the rates of change of the surface areas of the cube and sphere, given that their edge and radius increase at the same rate.
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