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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 31-60 of 72 questions
31 Easy
31 of 72
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.
32 Easy
32 of 72
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.
33 Easy
33 of 72
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.
34 Easy
34 of 72
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.
35 Easy
35 of 72
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.
36 Easy
36 of 72
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.
37 Easy
37 of 72
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.
38 Easy
38 of 72
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}}.
39 Easy
39 of 72
Application +4 | -1
A pointon the parabola y2=18xy^2=18 x at which the ordinate increases at twice the rate of the abscissa, is
View answer and explanation
Correct option: A
(9/8,9/2)(9 / 8,9 / 2)
Option A ((9/8,9/2)(9 / 8,9 / 2)) is correct because it satisfies the given condition that the ordinate increases at twice the rate of the abscissa on the parabola.
40 Easy
40 of 72
Application +4 | -1
For what values of xx is the rate of increase of x3−5x2+5x+8x^3-5 x^2+5 x+8 is twice the rate of increase of xx ?
View answer and explanation
Correct option: D
3,133, \frac{1}{3}
The correct answer is D (3,133, \frac{1}{3}) because these are the values of x that satisfy the condition where the rate of increase of the given function is twice the rate of increase of x.
41 Easy
41 of 72
Application +4 | -1
If a particle moving along a line follows the law s=1+ts=\sqrt{1+t}, then the accelertion is proportional to
View answer and explanation
Correct option: C
cube of the velocity
The correct answer is cube of the velocity because the acceleration is proportional to the cube of the velocity based on the given displacement equation.
42 Easy
42 of 72
Analyzing +4 | -1
A right circular cylinder which is open at the top and has a given surface area, will have the greatest volume if its height hh and radius rr are related by
View answer and explanation
Correct option: D
h=rh = r
Option D (h=rh = r) is correct because it represents the relationship between height and radius that maximizes the volume of the cylinder for a given surface area, found through differentiation.
43 Easy
43 of 72
Application +4 | -1
The weight WW of a certain stock of fish is given by W=nwW=n w, where nn is the size of stock and ww is the average weight of a fish. If nn and ww change with time tt as n=2t2+3n=2 t^2+3 and w=t2−t+2w=t^2-t+2, then the rate of change of WW with respect to t\mathrm{t} at t=1t=1 is
View answer and explanation
Correct option: C
13
Option C (13) is correct because the rate of change of the weight WW with respect to time tt at t=1t=1 is calculated to be 13.
44 Easy
44 of 72
Application +4 | -1
The coordinates of the point on the ellipse 16x2+9y2=40016 x^2+9 y^2=400 where the ordinate decreases at the same rate at which the abscissa increases, are
View answer and explanation
Correct option: A
(3,16/3)(3,16 / 3)
The correct answer is A ((3,16/3)(3,16 / 3)) because these coordinates satisfy the ellipse equation and the condition where the ordinate decreases at the same rate as the abscissa increases.
45 Easy
45 of 72
Application +4 | -1
The speed vv of a particle moving along a straight line is given by a+bv2=x2a+b v^2=x^2 where xx is its distance from the origin. The acceleration of the particle is
View answer and explanation
Correct option: A
xb\frac{x}{b}
Option B (103 cm2/sec10 \sqrt{3} \mathrm{~cm}^2 / \mathrm{sec}) is correct because it is the rate of increase of the area when each side is 10 cm.
46 Easy
46 of 72
Application +4 | -1
If x=−1x = - 1 and x=2x = 2 are extreme points of f(x)=αlog⁡∣x∣+βx2+x, f(x) = \alpha \log \left| x \right| + \beta {x^2} + x,\, then
View answer and explanation
Correct option: C
α=2,  β=−12\alpha = 2,\,\,\beta = - \frac{1}{2}
Option C (α=2, β=−12\alpha = 2,\,\beta = - \frac{1}{2}) is correct because substituting the extreme points into the derivative of the function and solving the resulting equations gives these values.
47 Easy
47 of 72
Application +4 | -1
The radius of a sphere is changing at the rate of 0.1 cm/sec0.1 \mathrm{~cm} / \mathrm{sec}. The rate of change of its surface area when the radius is 200 cm200 \mathrm{~cm} is
View answer and explanation
Correct option: C
160πcm2/sec160 \pi \mathrm{cm}^2 / \mathrm{sec}
Option C (160πcm2/sec160 \pi \mathrm{cm}^2 / \mathrm{sec}) is correct because it is the rate of change of the surface area of the sphere, calculated using related rates.
48 Easy
48 of 72
Application +4 | -1
While measuring the side of an equilateral triangle an error of k%k\% is made, the percentage error in its area is
View answer and explanation
Correct option: B
2k %2k\,\%
Option B (2k %2k\,\%) is correct because the percentage error in the area is twice the percentage error in the side length.
49 Easy
49 of 72
Application +4 | -1
If the velocity of a body moving in a straight line is proportional to the square root of the distance traversed, then it moves with
View answer and explanation
Correct option: B
constant force
The correct answer is constant force because if velocity is proportional to the square root of distance, the acceleration is constant, implying a constant force.
50 Easy
50 of 72
Application +4 | -1
If the surface area of a sphere of radius rr is increasing uniformly at the rate 8 cm2/s8 \mathrm{~cm}^2 / \mathrm{s}, then the rate of change of its volume is :
View answer and explanation
Correct option: D
proportional to rr
Option D (proportional to rr) is correct because the rate of change of volume is directly proportional to the radius rr of the sphere.
51 Easy
51 of 72
Application +4 | -1
If the rate of change of area of a circle is equal to the rate of change of its diameter, then its radius is equal to
View answer and explanation
Correct option: A
2π\frac{2}{\pi} unit
Option A (xb\frac{x}{b}) is correct because it is the acceleration derived from the given speed equation.
52 Easy
52 of 72
Application +4 | -1
Consider a rectangle whose length is increasing at the uniform rate of 2 m/sec2 \mathrm{~m} / \mathrm{sec}, breadth is decreasing at the uniform rate of 3 m/sec3 \mathrm{~m} / \mathrm{sec} and the area is decreasing at the uniform rate of 5 m2/sec5 \mathrm{~m}^2 / \mathrm{sec}. If after some time the breadth of the rectangle is 2 m2 \mathrm{~m} then the length of the rectangle is
View answer and explanation
Correct option: D
3 m3 \mathrm{~m}
Option D (3 m3 \text{{ m}}) is correct because the length of the rectangle is calculated to be 3 meters when the breadth is 2 meters.
53 Easy
53 of 72
Application +4 | -1
A particle moves on a line according to the law s=at2+bt+cs = a{t^2} + bt + c. If the displacement after 1 sec is 16 cm, the velocity after 2 sec is 24 cm/sec and acceleration is 8cm/sec⁡2  8\mathrm{cm}/{\sec ^2}\mathrm{\;}⁡ , then
View answer and explanation
Correct option: A
a=4,  b=8,  c=4a = 4,\;b = 8,\;c = 4
Option A (a=4,  b=8,  c=4a = 4,\;b = 8,\;c = 4) is correct because these values satisfy the given conditions for displacement, velocity, and acceleration.
54 Easy
54 of 72
Application +4 | -1
If an error of 1∘1^{\circ} is made in measuring the angle of a sector of radius 30 cm30 \mathrm{~cm}, then the approximate error in its area, is
View answer and explanation
Correct option: C
2.5πcm22.5 \pi \mathrm{cm}^2
Option C (2.5πcm22.5 \pi \mathrm{{cm}}^2) is correct because it is the approximate error in the area of the sector calculated using differentials.
55 Easy
55 of 72
Analyzing +4 | -1
A particle moves along the X- axis as x=u(t−2)+a(t−2)2x = u\left( {t - 2} \right) + a{\left( {t - 2} \right)^2}

(i) The initial velocity of particle is uu

(ii) The acceleration of particle is aa

(iii) The acceleration of the particle is 2a

(iv) At t=2s , particle is at the origin

Among these correct statements are
View answer and explanation
Correct option: C
(iii), (iv)
Option C ((iii), (iv)) is correct because the acceleration is 2a and at t=2s, the particle is at the origin.
56 Easy
56 of 72
Application +4 | -1
If an error of k%k\% is made in measuring the radius of a sphere, then percentage error in its volume is
View answer and explanation
Correct option: B
3k%3k\%
Option B (3k%3k\%) is correct because the percentage error in volume is three times the percentage error in the radius of the sphere.
57 Easy
57 of 72
Application +4 | -1
Spherical rain drop evaporates at a rate proportional to its surface area. The differential equation corresponding to the rate of change of the radius of the rain drop, if the constant of proportionality is K>0K>0, is
View answer and explanation
Correct option: A
drdt+K=0\frac{d r}{d t}+K=0
Option A, drdt+K=0\frac{d r}{d t}+K=0, is the correct differential equation representing the rate of change of the radius, where the rate of evaporation is proportional to the surface area.
58 Easy
58 of 72
Application +4 | -1
A lizard, at an initial distance of 21 cm21 \mathrm{~cm} behind an insect, moves from rest with an acceleration of 2 cm/s22 \mathrm{~cm} / \mathrm{s}^2 and pursues the insect which is crawling uniformly along a straight line at a speed of 20 cm/s20 \mathrm{~cm} / \mathrm{s}. Then the lizard will catch the insect after
View answer and explanation
Correct option: C
21 s21 \mathrm{~s}
Option C (21 s21 \text{{ s}}) is correct because the time it takes for the lizard to catch the insect is calculated to be 21 seconds.
59 Easy
59 of 72
Application +4 | -1
If the volume of a sphere is increasing at a constant rate, then the rate at which its radius is increasing, is
View answer and explanation
Correct option: D
Inversely proportional to the surface area
Option D (Inversely proportional to the surface area) is correct because the rate of change of the radius is inversely proportional to the square of the radius, which is proportional to the surface area.
60 Easy
60 of 72
Application +4 | -1
If V=43πr3V=\frac{4}{3} \pi r^3, at what rate in cubic units is VV increasing when r=10r=10 and drdt=0.01\frac{d r}{d t}=0.01 ?
View answer and explanation
Correct option: B
4π4 \pi
Option B (4π4 \pi) is correct because it is the rate of change of volume calculated using related rates, given the rate of change of the radius.
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