Free Learning Resources
PW Books
Notes (Class 10-12)
PW Study Materials
Notes (Class 6-9)
Ncert Solutions
Govt Exams
Class 6th to 12th Online Courses
Govt Job Exams Courses
UPSC Coaching
Defence Exam Coaching
Gate Exam Coaching
Other Exams
Know about Physics Wallah
Physics Wallah is an Indian edtech platform that provides accessible & comprehensive learning experiences to students from Class 6th to postgraduate level. We also provide extensive NCERT solutions, sample paper, NEET, JEE Mains, BITSAT previous year papers & more such resources to students. Physics Wallah also caters to over 3.5 million registered students and over 78 lakh+ Youtube subscribers with 4.8 rating on its app.
We Stand Out because
We provide students with intensive courses with India’s qualified & experienced faculties & mentors. PW strives to make the learning experience comprehensive and accessible for students of all sections of society. We believe in empowering every single student who couldn't dream of a good career in engineering and medical field earlier.
Our Key Focus Areas
Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.
What Makes Us Different
Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others
Copyright © 2025 Physicswallah Limited All rights reserved.
= 0.497
Therefore, the area is decreasing at the rate of √3b cm
2
/s.
Hence, the perpendicular distance of the normal from the origin is constant.
is (i) increasing (ii) decreasing.
Solution :
Question
7. Find the intervals in which the function f given by
is (i) increasing (ii) decreasing.
Solution :
Question
8. Find the maximum area of an isosceles triangle inscribed in the ellipse
with its vertex at one end of the major axis.
Solution :
Equation of the ellipse is
Let the major axis be along the
x
−axis.
Let ABC be the triangle inscribed in the ellipse where vertex C is at (
a
, 0).
Since the ellipse is symmetrical with respect to the
x
−axis and
y
−axis, we can assume the coordinates of A to be (−
x
1
,
y
1
) and the coordinates of B to be (−
x
1
, −
y
1
).
Question 9. A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs ` 70 per sq. meter for the base and ` 45 per square meter for sides. What is the cost of least expensive tank?
Solution :
Given: Depth of tank = 2 m
Let
l
,
b
, and
h
represent the length, breadth, and height of the tank respectively.
Then, we have height (
h)
= 2 m
Volume of the tank = 8m
3
Volume of the tank =
l
×
b
×
h
∴ 8 =
l
×
b
× 2
lb = 4, b = 4/l
Now, area of the base =
lb
= 4
Area of the 4 walls (
A)
= 2
h
(
l
+
b)
Thus, by second derivative test, the area is the minimum when
l
= 2.
We have
l
=
b
=
h
= 2.
∴Cost of building the base = Rs 70 × (
lb
) = Rs 70 (4) = Rs 280
Cost of building the walls = Rs 2
h
(
l
+
b
) × 45 = Rs 90 (2) (2 + 2)
= Rs 8 (90) = Rs 720
Required total cost = Rs (280 + 720) = Rs 1000
Hence, the total cost of the tank will be Rs 1000.
Question
10. The sum of the perimeter of a circle and square is k where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.
Solution :
Let r be the radius of the circle and a be the side of square.
Therefore, sum of areas is minimum when side of the square is double the radius of the circle.
Question
11. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.
Solution :
Let
x
and
y
be the length and breadth of the rectangular window.
Radius of the semicircular opening = x/2
It is given that the perimeter of the window is 10 m.
∴Area of the window (
A)
is given by,
Thus, when
Therefore, by second derivative test, the area is the maximum when length x = 20/π + 4m.
Hence, the required dimensions of the window to admit maximum light is given bylength = 20/π+4m and breadth = 10/π+4m.
Question
12. A point on the hypotenuse of a triangle is at distances a and b from the sides of the triangle. Show that the maximum length of the hypotenuse is
Solution :
Let ΔABC be right-angled at B. Let AB =
x
and BC =
y
.
Let P be a point on the hypotenuse of the triangle such that P is at a distance of
a
and
b
from the sides AB and BC respectively.
Let ∠C =
θ
.
We have,
AC = √x
2
+ y
2
Now,
PC =
b
cosec
θ
And, AP =
a
sec
θ
∴AC = AP + PC
⇒ AC =
b
cosec
θ
+
a
sec
θ
… (1)
Therefore, by second derivative test, the length of the hypotenuse is the maximum when tan θ(b/a)
1/3
Now, when tan θ(b/a)
1/3
, we have:
Hence, the maximum length of the hypotenuses is
Now, for values of
x
close to 2/7 and to the left of 2/7, f'(x) > 0. Also, for values of
x
close to 2/7 and to the right of 2/7, f'(x) < 0
Thus, x = 2/7 is the point of local maxima.
Now, for values of
x
close to 2 and to the left of 2, f'(x) < 0 Also, for values of
x
close to 2 and to the right of 2, f'(x) > 0
Thus,
x
= 2 is the point of local minima.
Now, as the value of
x
varies through −1,f'(x) does not changes its sign.
Thus,
x
= −1 is the point of inflexion.
Question
14. Find the absolute maximum and minimum values of the function f given by f(x) = cos
2
x + sin x , x ∊ [0,π ]
Solution :
Given:f(x) = cos
2
x + sin x ……….(i)
Now, evaluating the value of
f
at critical points x = π/2 and x = π/6 and at the end points of the interval [0,π ] (i.e., at
x
= 0 and
x
= π), we have:
Hence, the absolute maximum value of
f
is 5/4 occurring at x = π/6 and the absolute minimum value of
f
is 1 occurring at x = 0, π/2, and π.
Question
15. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3
Solution :
Let r be the radius of base of cone and h be the height of the cone inscribed in a sphere of radius
The volume (
V)
of the cone is given by,
V = 1/3πR
2
h
Now, from the right triangle BCD, we have:
BC = √r
2
- R
2
∴
h
= r+√r
2
- R
2
Hence, it can be seen that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius
r
is 4r/3.
Question
16. Let
f
be a function defined on [
a
,
b
] such that
f
'(
x
) > 0, for all
x
∈ (
a
,
b
). Then prove that
f
is an increasing function on (
a
,
b
).
Solution :
Let x1, x2∈(a,b) such that x1<x2.
Consider the sub-interval [x1, x2]. Since f(x) is differentiable on (a, b) and [x1, x2]⊂(a,b).
Therefore, f(x) is continous on [x1, x2] and differentiable on (x1, x2).
By the Lagrange's mean value theorm, there exists c∈(x1, x2) such that
f'(c)=f(x2)-f(x1)x2-x1 ...(1)
Since f'(x) ) > 0 for all x∈(a,b), so in particular, f'(c) > 0.
f'(c)>0⇒f(x2)-f(x1)x2-x1>0 [Using (1)]
⇒f(x2)-f(x1)>0 [∵ x2-x1>0 when x1<x2]
⇒f(x2)>f(x1)⇒f(x1)<f(x2)
Since x1, x2 are arbitrary points in (a,b).
Therefore, x1<x2⇒f(x1)<f(x2) for all x1,x2∈(a, b)
Hence,f(x) is increasing on (a,b).
Question
17. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3 Also find the maximum volume.
Solution :
A sphere of fixed radius (
R)
is given.
Let
r
and
h
be the radius and the height of the cylinder respectively.
From the given figure, we have h = 2√R
2
- r
2
The volume (
V)
of the cylinder is given by,
Now, it can be observed that at
∴The volume is the maximum when r
2
= 2R
2
/3
When r
2
= 2R
2
/3, the height of the cylinder is
Hence, the volume of the cylinder is the maximum when the height of the cylinder is
2R/√3
Question
18. Show that the height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and having semi-vertical angle
α
is one-third that of the cone and the greatest volume of the cylinder is 4/27πh
3
tan2
α
.
Solution :
The given right circular cone of fixed height (
h)
and semi-vertical angle (
α
)
can be drawn as:
Here, a cylinder of radius
R
and height
H
is inscribed in the cone.
Then, ∠GAO =
α
, OG =
r
, OA =
h
, OE =
R
, and CE =
H
.
We have,
r
=
h
tan
α
Now, since ΔAOG is similar to ΔCEG, we have:
Now, the volume (
V)
of the cylinder is given by,
∴By second derivative test, the volume of the cylinder is the greatest when
Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.
Now, the maximum volume of the cylinder can be obtained as:
Choose the correct answer in the Exercises 19 to 24:
Question
19. A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic meter per hour. Then the depth of wheat is increasing at the rate of:
(A) 1 m/h
(B) 0.1 m/h
(C) 1.1 m/h
(D) 0.5 m/h
Solution :
Let
r
be the radius of the cylinder.
Then, volume (
V)
of the cylinder is given by,
Differentiating with respect to time
t
, we have:
The tank is being filled with wheat at the rate of 314 cubic metres per hour.
Thus, we have:
Hence, the depth of wheat is increasing at the rate of 1 m/h.
The correct answer is A.
Question
20. The slope of the tangent to the curve x = t
2
+ 3t – 8, y = 2t
2
– 2t – 5 at the point (2,– 1) is
(A) 22/7
(B) 6/7
(C) 7/6
(D) -6/7
Solution :
The given curve is x = t
2
+ 3t – 8 and y = 2t
2
– 2t – 5
The given point is (2, −1).
At
x
= 2, we have:
The common value of
t
is 2.
Hence, the slope of the tangent to the given curve at point (2, −1) is
Therefore, option (B) is correct.
Question
21.
The line
y
=
mx
+ 1 is a tangent to the curve
y
2
= 4
x
if the value of
m
is
(A) 1
(B) 2
(C) 3
(D) 1/2
Solution :
The equation of the tangent to the given curve is
y
=
mx
+ 1.
Now, substituting
y
=
mx
+ 1 in
y
2
= 4
x
, we get:
Since a tangent touches the curve at one point, the roots of equation (i) must be equal.
Therefore, we have:
Discriminant = 0
Hence, the required value of
m
is 1.
Therefore, option (A) is correct.
Question
22.
The normal at the point (1, 1) on the curve 2
y
+
x
2
= 3 is
The slope of the normal to the given curve at point (1, 1) is
Hence, the equation of the normal to the given curve at (1, 1) is given as:
⇒ y - 1 = 1(x - 1)
⇒ y - 1 = x - 1
⇒ x - y = 0
Therefore, option (B) is correct.
Question
23.
The normal to the curve
x
2
= 4
y
passing (1, 2) is
The slope of the normal to the given curve at point (
h
,
k
) is given by,
∴Equation of the normal at point (
h
,
k
) is given as:
Now, it is given that the normal passes through the point (1, 2).
Therefore, we have:
Since (
h
,
k
) lies on the curve
x
2
= 4
y
, we have
h
2
= 4
k.
⇒ k =h
2
/4
From equation (i), we have:
Hence, the equation of the normal is given as:
Therefore, option (A) is correct.
Question
24. The points on the curve 9
y
2
=
x
3
where the normal to the curve make equal intercepts with axes are:
Solution :
The equation of the given curve is 9
y
2
=
x
3
.
Differentiating with respect to
x
, we have:
The slope of the normal to the given curve at point (x
1
,y
1
) is
∴ The equation of the normal to the curve at (x
1
,y
1
) is
It is given that the normal makes equal intercepts with the axes.
Therefore, We have:
Also, the point(x
1
,y
1
)lies on the curve, so we have
9y
1
2
= x
1
3
.....................(ii)
From (i) and (ii), we have:
Therefore, option (A) is correct.
