1
SANGHARSH ASSIGNMENT
Basic Mathematics
1.
Find value of given mathematical expression.
−
(i)
e
(iii)
1
0.001
(ii) cos 3°
6
(iv)
0.3
(v)
0.49
(vi) e0
8.
If a triangle ABC, the side AB is 8 cm, side BC is
8 cm, and side AC is 8 2 cm. What is the measure
of angle C?
(1) 30°
(2) 45°
(3) 60°
(4) 90°
(vii) 1 − 0.19
2.
Find area of given graph
Assignment-04
By: M.R. Sir
9.
If cos () =
1
2
and cos () =
3
, where 0° < ,
2
 < 90°, find the value of sin ( + ).
(1) 0.58
(2) 0.72
(3) 1.85
(4) 0.96
10.
3.
4.
5.
6.
7.
Which of the following formula is wrong.
(1) sin 2 = 2 sin  . cos 
(2) cos (2) = cos2 – sin2


(3) sin  = 2sin cos
2
2


(4) cos  = sin 2 − cos 2
2
2
Find distance between centre to corner of equilateral
triangle of side a.
Find value of different trigonometric ratio
(i) sin (–45°)
(ii) cos (405°)
(iii) sin (390°)
(iv) sin (300)
(v) tan (–120°)
If sin () = 0.6 and cos () = 0.8, where  and  are
acute angles, what is the value of sin ( + )?
(1) 0.28
(2) 0.48
(3) 0.96
(4) 1.88
In a right triangle, the length of the hypotenuse is
10 cm and one of the acute angles is 30°. What is
the length of the side opposite to the 30° angle?
(1) 5 cm
(2) 5 3 cm
(3) 10 cm
(4) 10 3 cm
Find value of x for following expression:
1
(i)
= 10− x
0.001
(ii)
3x = 1
(iii)
0.49 = 700  10− x
(iv)
1 − x = 33/2
(v)
0.125
= e
(1 / x)
(vi)
6
= 5.(2 x )
0.3
x
153  162  13  3
= (9) 2
(vii)
18  17  117
(viii)
(ix)
(x)
(4 x + 4−1 )  22
125
=
 10−2
10
5
2
11.
Correct matrix match
16.
(A) I = sin (t – /4)
(i)
V = cos (t – 3/4)
(B) I = sin (t + /6)
V = cos (t + /3)
(C) I = sin (t – /6)
V = cos (t + /6)
(D) I = sin (t + 2/3)
V = cos (t – /6)
Current
lay
voltage by 2/3
Find x in given expression:
(i)
(ii) Voltage leads by
current by /6
(ii)
(iv)
(iv) Voltage is in
phase of current
4
4
2
( 2)
x
= 4x
0.001 10−4
10  10
−3
= 10 x
(vi) y 2 . y 3/2 = y x
Correct matrix match
4
5
4200  0.001
(A) sin (–120)
(1)
(B) cos (150)
(2) −
(C) tan (135)
3
(3) −
2
(ix) 0.52  10−8 = 520  10 x
(D) tan (143)
(4) –1
(E) sin (127)
3
(5) −
4
0.75  30
= 4x
45  0.25
(vii)
3
2
(viii)
(x)
If 64 identical liquid sphere combine to bigger
sphere then find radius of bigger sphere.
14.
A wire of length L bend into circle and square then
find ratio of radius of circle and side of square.
15.
Which of the following is True/False.
(i) 23 × 22 = 26
(ii) 210 × 310 = 510
(iii) 210 × 310 = 610
(iv)
xy = x  y
(v)
x+ y = x + y
2
2
(vi) 4 + 3 = 7
2
4 3

3 4
2 7

(ix)
3 5
(x) 2.0 = 0.02 × 10–2
(viii)
1 1
(vii) 
3
3
10
−2
0.326  10−4
10
+3
= 42  10 x
= 32.6  10 x
1
17.
13.
1
27 = 3 x
(iii) 32 =
(iii) Current leads by
voltage /3
(v)
12.
1 1 1
+ + =x
2 5 7
18.
19.
In a decay process the number of nuclei N = N0e–t
where N0 and  is constant, find rate of decay
(1) N0
(2) –N
(3) N0e–t
(4)

N0
Focal length ‘f ’ of a lens varies with temperature as
df
f = f 0 (1 + T ) find
dT
(1) f0
(2) f0
(3) 
(4) f0/
Electric potential in a conservative field varies with
space as V = ax2 + by + cz, find electric field at point
dV
(1, 1) if E = −
dr
(1) 2aiˆ + bjˆ
(2) 2iˆ − bjˆ
(3) 2aiˆ − bjˆ
(4) −(2aiˆ + bjˆ)
3
20.
21.
 = a cos t , x = a (t − sin t ) find
(1) –1
(2) 1
(3) 0
(4)
x = a cos3 t ,  = a sin 3 t find
(1) cot t
22.
(3)
1
2
(3)
y − x2
2
y −x
2
2
2
y +x
(1) 2 + 6t
24.
25.
A
T
(2) −
A
T2
(4) zero
(4)
y2 − x
y − x2
y+x
2
(1) 45°
(2) 60°
(3) 37°
(4) 53°
2
y +x
d2y
dx 2
(2) 2 – 6t
2
u 2 sin 2
as shown below, find angle of
g
(4) 2 – 6t2
A charged particle is released in a uniform
horizontal electric field. At what initial angle it
should be projected so that maximum height
attained, if maximum height attained is directly
proportional to vertical velocity and force inside
electric field is F = qE
A charged particle moves such that x(t) = Aekt what
is the electric current if charge is proportional to x.
(1) kAekt
(2) Aekt
(3) Ake–kt
(4) zero
(1)  = 0°
(2)  = 90°
(3)  = 180°
(4)  = 45°
3
Kinetic energy of a particle is given as k = at , find
power delivered to particle if power is rate of
change in kinetic energy w.r.t. time:
(1) 3at2
(3) at
26.
T
2
projection for which range will be maximum.
x = t + 1 and y = t2 + t3 find
(3) 2 + 6t
A
Horizontal range in projectile motion can be given
as
dy
dx
(2)
y2 − x
28.
1
cot t
29.
23.
Volume of a gas changes with temperature as
A
V = , find rate of change in volume w.r.t.
T
temperature. [A = constant]
(1)
dy
dx
(4) −
x3 + y3 = 3xy, find
(1)
27.
(2) tan t
1
tan t
(3) −
dy

at t =
2
dx
2
(2) 2at
30.
In prism experiment deviation can be given as
 = i + e – A as shown below. Then find the
condition if deviation is minimum.
(4) None
A particle moves such that its velocity and position
are related by v = kx, if acceleration is defined as
v dv
find its acceleration at x = 2m.
dx
(1) i = 0°
(2) i = 90°
(1) 2k
(2) 2k2
(3) e = 0°
(4) i = e
(3) 4k
(4) zero
4
31.
32.
The potential energy stored in a spring varies as
1
U = kx 2 + x3 , for what value of x, magnitude of
2
force is maximum if force and potential energy
U
related as F = −
x
(1)
k
x=
6
6k
(2) x =

(3)
−6k
x=

−k
(4) x =
6
Work done by gas is given by dW = P.dV, if
V = ax2 – bx, find value of x at which work is
maximum.
(1)
a
2b
(2)
(3)
a
b
(4) −
b
2a
a
b
33.
Force on charge ‘q’ on axial point at a distance ‘x’
from centre of uniformly charged ring is
kqx
where, k is constant, R is radius of
2
( R + x 2 )3/2
ring. For what value of x, force will be maximum.
(1) x =
(3)
R
2 2
2R
(2) x =
(4) R
R
2
5
ANSWER KEY
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
(i) 0
(iii) 103
(v) 0.7
(vii) 0.9
(30)
(4)
a
(ii) 1
(iv) 20
(vi) 1
3
(i)
−
(iii)
1
2
1
(ii)
2
2
1
=
2
2
(iv) −
3
2
3
(v)
(3)
(1)
(2)
(4)
(i) –3
(ii) 0
(iii) 3
(iv) –26
(v) 8
(vi) 2
(vii) 3
(viii) 0
(ix) 30
(x) 3
(A) → (iv)
(B) → (i)
(C) → doesn’t match any option
(D) → (iii)
12.
(A) → (2) –
13.
14.
15.
(B) → (3)
(C) → (4)
(D) → (5)
(E) → (1)
R = 4r
2/
(i)
False
3/2
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
(ii) False
(iii) True
(iv) True
(v)
False
(vi) False
(vii) True
(viii) False
(ix) True
(x)
False
(i)
59/70
(ii) 1/3
(iii) 10
(iv) –3/2
(v)
–5
(vi) 7/2
(vii) x = 1
(viii) –9
(ix) –11
(x) x = 2
(2)
(1)
(4)
(1)
(4)
(1)
(1)
(1)
(1)
(2)
(2)
(1)
(2)
(4)
(4)
(2)
(2)
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