Skip to main content
Back to RATE MEASURE

Your Progress

0 / 110 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • RATE MEASURE

Rate measure practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 61-90 of 110 questions

Preparing practice and saved progress…

61 Medium
61 of 110
Application +4 | -1
If the radius of a circle be increasing at a uniform rate of 2cm/s. The area of increasing of area of circle ,at the instant when the radius is 20 cm, is
View answer and explanation
Correct option: C
80πcm2/s  80\pi c{m^2}/s\;
Option C (80πcm2/s  80\pi c{m^2}/s\;) is correct because it represents the rate of increase of the area of the circle when the radius is 20 cm, calculated using related rates.
62 Medium
62 of 110
Application +4 | -1
The distance ss meters covered by a body in t seconds,is given by s=3t2−8t+5.s = 3{t^2} - 8t + 5. The body will stop after
View answer and explanation
Correct option: C
43s\frac{4}{3}s
Option C (43s\frac{4}{3}s) is correct because it represents the time at which the velocity of the body becomes zero, indicating that the body has stopped.
63 Easy
63 of 110
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.
64 Medium
64 of 110
Application +4 | -1
If a particle moves such that the displacement is proportional to the square of the velocity acquired , then its acceleration is
View answer and explanation
Correct option: D
A constant
Option D (A constant) is correct because if displacement is proportional to the square of the velocity, then the acceleration is constant.
65 Easy
65 of 110
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.
66 Easy
66 of 110
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.
67 Medium
67 of 110
Application +4 | -1
The pressure PP and volume VV of a gas are connected by the relation PV1/4=constant.P{V^{1/4}} = \mathrm{constant}. The percentage increase in the pressure corresponding to a deminition of 1/2  %1/2\;\% in the volume is
View answer and explanation
Correct option: C
18%\frac{1}{8}\%
Option C (18%\frac{1}{8}\%) is correct because it represents the percentage increase in pressure corresponding to a diminution of 1/2 % in volume, calculated using the given relation.
68 Easy
68 of 110
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.
69 Easy
69 of 110
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.
70 Easy
70 of 110
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.
71 Medium
71 of 110
Application +4 | -1
A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of 50cm3/min⁡\mathrm{50 c}{\mathrm{m}^\mathrm{3}}/\min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of the ice decreases, is
View answer and explanation
Correct option: B
118π\frac{1}{{18\pi }}
Option B (118π\frac{1}{{18\pi }}) is correct because it represents the rate at which the thickness of the ice decreases, calculated using related rates and the volume of a sphere.
72 Easy
72 of 110
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.
73 Medium
73 of 110
Application +4 | -1
The maximum area of the rectangle that can be inscribed in a circle of radius rr, is
View answer and explanation
Correct option: D
2r22{r^2}
Option D (2r22{r^2}) is correct because it represents the maximum area of a rectangle inscribed in a circle, derived through calculus.
74 Easy
74 of 110
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.
75 Easy
75 of 110
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.
76 Easy
76 of 110
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.
77 Medium
77 of 110
Application +4 | -1
A missile is fired from the ground level rises xx metres vertically upwards in tt seconds where x=100t−252t2.x = 100t - \frac{{25}}{2}{t^2}. The maximum height reached is
View answer and explanation
Correct option: A
200 m
Option A (200 m) is correct because it is the maximum height reached by the missile, calculated by finding when the velocity is zero and substituting that time back into the height equation.
78 Medium
78 of 110
Application +4 | -1
The distance travelled by a particle upto time xx is given by f(x)=x3−2x+1.f\left( x \right) = {x^3} - 2x + 1. The time cc at which the velocity of the particle is equal to its average velocity between times x=1x = 1 sec and x=2x = 2 sec, is
View answer and explanation
Correct option: C
3  sec\sqrt 3 \;\mathrm{sec}
Option C (3  sec\sqrt 3 \;\mathrm{sec}) is correct because it represents the time at which the instantaneous velocity equals the average velocity over the given interval, as determined by the Mean Value Theorem.
79 Easy
79 of 110
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.
80 Easy
80 of 110
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.
81 Easy
81 of 110
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.
82 Easy
82 of 110
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.
83 Easy
83 of 110
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.
84 Easy
84 of 110
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.
85 Easy
85 of 110
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.
86 Easy
86 of 110
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.
87 Easy
87 of 110
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.
88 Easy
88 of 110
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.
89 Easy
89 of 110
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.
90 Easy
90 of 110
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.
Revision notes and key takeaways

Review RATE MEASURE in your textbook or class notes, then use this set to check your understanding.

Showing 61-90 of 110 questions

Your practice

0 / 110 answered · 110 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.