Solutions for Ultimate JEE Main Physics Grand Test-I

1. Let the mass is represented by M then M = f (V, F, T)

Assuming that a function is product of power functions of V, F and T
M = KVx Fy Tz
Where K is a dimension less constant of proportionality. The above equation dimensionally becomes.
[M] = [LT–1]x [MLT–2]y [T]2
i.e. [M] = [My] [Lx + y T – x – 2y + z ]
So equation becomes
[M] = [My Lx + y T– x – 2y + z]
For dimensionally correct expression,
y = 1, x + y = 0 and – x – 2y + z = 0
 x = –1, y = 1 and z = 1.
therefore M = KV–1 FT.
Hence correct answer is (B).
2. As shown in fig for quarter revolution
  
v = v 2 − v1 and  = 90º,

So v = v 2 + v 2 = ( 2 )v
v
 = tan–   = 45º
v

v = 2v south west.
North
v1
West v2
East

–v1
v

South
Hence correct answer is (A).

3. Let x be the distance travelled by the truck when truck and car are side by side. The distance travelled by
the car will be (x + 150) as the car is 150 metre behind the truck. Applying the formula
s = ut + (1/2) a t2,
we have, x = 1/2 × (1.5) t2 ........(1)
and (x + 150) = (1/2) × (2) t2 ........(2)
Here t is the common time.
From equations (1) and (2)
x + 150 2
we have, = Solving we get, x = 450 m (truck) and x + 150 = 600 m (car)
x 1 .5
Substituting the value of x in eq. (1), we get
450 = 1/2 (1.5) t2
VECTOR TIRUPATI.CELL NO.9440025125.
 450  2
t= = 600 = 24.5 sec.
1.5
Hence correct answer is (C)

4. Let ux and uy be the components of the velocity of the particle along the x- and y-directions. Then
ux = dx/dt = u0 and uy = dy/dt = a cos t
Integration : x = u0 t and y = a sin t
Eliminating t : y = a sin (x/u0)
This is the equation of the trajectory
At t = 3/2, we have, x = u0 3/2 and
y = a sin 3/2 = – a
 3u  2 
 The displacement of the particle from the origin is x +y =
2 2
 0
 + a2
 2  

Hence correct answer is (A)

5. As P and Q move down, the length l decreases at the rate of U m/s

b b
A B
y
 

O Q
P

M
From figure, l2 = b2 + y2
Differentiating with respect to time
d dy
2l = 2y (b is constant)
dt dt
dy  d 1 d U
 = . = . =
dt y dt cos  dt cos 
Hence correct answer is (D)

6. The following two forces are acting on the body

(i) Weight mg is acting vertically downward
(ii) The push of the air is acting upward.
As the body is accelerating downward, the resultant force is (mg – F)
Workdone by the resultant force to fall through a vertical
distance of 20 m = (mg – F) × 20 joule
1
Gain in the kinetic energy = mv2
2
Now the workdone by the resultant force is equal to the change in kinetic energy i.e.
1 1
(mg – F) 20 = mv2 (From work-energy theorem) or (50 – F) 20 = × 5 × (10)2 or 50 – F =
2 2
12.5
or F = 50 – 12.5  F = 37.5 N
Work done by the force
= – 37.5 x 20 = – 750 joule

VECTOR TIRUPATI.CELL NO.9440025125.

(The negative sign is used because the push of the air is upwards while the displacement is downwards.)

7.
O



T cos 
T T

A T sin 

mg mg
From figure
T cos  = mg ....(1)
2 2
mv mv
T sin  = = .... (2)
r  sin
mg
From eq. (1) T =
cos 
When the string is horizontal,  must be 90º i.e., cos 90º = 0
mg
 T= =
0
Thus the tension must be infinite which is impossible, so the string can not be in horizontal plane.
The maximum angle  is given by the
breaking tension of the string in the equation
T cos  = m.g
Here T (Maximum) = 8 N and m = 0.4 kg
 8 cos  = 0.4 × g = 0.4 × 10 = 4
1
cos  = (4/8) = ,  = 60º
2
The angle with horizontal = 90º – 60º = 30º
0 .4  v 2
From equation (2), 8 sin 60º =
4 sin 60 o
32 sin 2 60 º
v2 = = 80 sin2 60º
0.4
 v = 80 sin 60º = 7.7 m/sec
Hence correct answer is (A)

8. The P.E of the mass at d/2 due to the earth and moon is
Earth Moon
R1 P O2
O1 m R2
M2
M1
d
GM1m GM 2 m
U=–2 –2
d d

VECTOR TIRUPATI.CELL NO.9440025125.

 2Gm
or U = – (M1 + M2) (Numerically)
d
1
m Ve2 = U
2
G
 Ve = 2 ( M1 + M 2 )
d

k
1
4
k 2
9. m 3

2k 2k
Effective constant of spring (B) & (C) = k '
1 1 1 1 2
 = +  = k'=k
k 2k 2k k 2k
k 1
4
k
m 5

k
Effective constant of spring (A) & (5) = k''
k'' = k + k = 2k
4
k 2k 6
m

Effective constant of spring (D) & (6) = k"'

 k"' = k + 2k = 3k
m
 Time period = T = 2
k
m
T = 2
3k
Hence correct answer is (D)
10. [B]
Change in area
A = 4 (nr2 – R2)
 energy evolved E
TA = 4T (nr2 – R2)
But nr3 = R3
 R3  1 1 
 E = 4T  3 r 2 − R 2  = 4TR3  − 
r  r R  
1
According to question E = Mv2
2
1 1  1 4
 4TR3  −  = × R3 dv2.
r R 2 3

6T  1 1 
 v=  − 
d r R
11. [C]
Force due to surface tension = weight of wire
2Tl = mg

2Tl = volume × density × g

VECTOR TIRUPATI.CELL NO.9440025125.

 2Tl = r2 ldg
2T
 r2 =
dg

2T
 r= ....(1)
dg

According to question T = 0.045 N/m

d = 8.96 × 103 kg/m3.
g = 98 m/s2.
From eqs. (A) and (B)
2  0.045
r=
3.14  8.96  10 3  9.8

r = 0.6 × 10-3 m
r = 0.6mm
diameter = 2r = 1.2 mm
12. Initially the position of wooden block is as shown in Fig.(a). Since the density of block is half than that of
water, hence half of its volume is immersed in water.
W
/2
/2


(a) (b)
When weight W is put on the block, the remaining half of the volume of block is immersed in water, Fig.(b).
Therefore,
W = additional upthrust + spring force
 
=×× × 2 × g + k  
2 2
 k
=    2g + 
 2

13. The equation of a plane progressive wave moving in the + x direction is

 t x
y = a sin 2  − , .....(1)
T 
Where a is the amplitude, T is the time period, &  is the wavelength.
Here a = 2m, v = 45m/sec n = 75 sec–1.
 T = 1/n = 1/75 sec.
45
and  = v/n = = 0.6 meter = 60 cm
75
Put these values of a, T &  in the above equation(1)
 x 
y = 2 sin 2  75t − ,
 60 
where y and x are in cm and t is in second.
To find out the displacement at a distance 135 cm from the origin at the instant t = 3sec, we substitute t = 3
sec. and x = 135 cm in the above equation
Therefore y = 2 sin 2 (225 – 2.25)
y = 2 sin (450 – 4.5)

VECTOR TIRUPATI.CELL NO.9440025125.

 y = + 2 sin (–4.5)
y = – 2sin (+ 4.5 )
y = – 2 sin(4 + /2)
y = – 2sin /2 = –2cm
14. The work done in a thermodynamic process is equal to the area enclosed between the
P-V curve and the volume axis.
work done by the gas in the process A → B is
W1 = area ABB'A' = AB × A' × A
 W1 = (6.0 – 1.0) litre × (12 × 105) Nxm2

 W1 = 5.0 × 10-3 m3 × 12 × 105 N/m2

 W1 = 6000 N-m = 6000J
work done in the process B → C is zero since volume remains constant
work done on the gas in the procces C → D is
W2 = area DCB'A'

W2 = DC × AD' = (5 × 10–3) × (2 × 105) = 1000J

work done in the process D → A is also zero
Hence the correct answer is (A)
VC rC

15. (B) 
VA

dV = – E  dr
rA

  
VC – VA = – E ( rC – rA )
 
= E  rCA
= ErCA cos 
= Er cos 0°
= Er

16. (C)
Let the current, in upper branch is I1 and in lower branch I2. The current in central resistance will be
I1 + I2. Using Kirchhoff's laws. 2 = I1 (2) + (I1 + I2) (2) upper branch
2 = I2 (2) + (I1 + I2) (2) lower branch
adding 4 = 2(I1 + I2) + 4(I1 + I2)
or I1 + I2 = 4/6 = 2/3 ampere. = 0.67 ampere
17. (A)
B0 = BPSR + BPQR .... (A)
µ i  2 − 2  µi
BPSR = 0   = 0 [ − ]
2  r  2 r
.... (B)
µ0i 2 sin 
BPQR = .
4 OQ
From the figure OQ = r cos 
µ i 2 tan 
BPQR = 0 .... (C)
4 r
From eqs. (A) and (C)
µ i µ i
B = 0 [ − ] + 0 tan 
2 r 2 r

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 µ 0i
= [ –  + tan ]
2 r
× ×
18. (D) × × × ×
× × × × × ×
B × × ×
× × a dr
× × × × × ×
× × × ×b × ×
× × × × × ×
The induced emf is obtained by considering a strip on the disc fig. Then, the linear speed of a small element
dr at a distance r from the centre is = r. The induced emf across the ends of the small element is-
de = B(dr)v = B r dr
Thus the induced emf across the inner and outer sides of the disc is
b
 Br dr =
1
e= B (b2 – a2)
a 2

19. (B)
Initially, the current lags behind the potential difference. Hence the circuit contains resistance and
inductance. The power of the circuit is –
P = Vrms × irms × cos .
Vrms
But irms =
Z
where Z = [R 2 + (L)2 ] is the impedance of the circuit.
Vrms
 P = Vrms × × cos .
Z
(V )2  cos  (220)2  0.8
or Z = rms =
P 550
= 70.4 ohm.
R
Power factor, cos  =
Z
 R = Z cos  = 70.4 × 0.8 = 56.32 ohm.
Now Z2 = R2 + (L)2
 (L)2 = Z2 – R2 = (70.4)2 – (56.4)2 = 1784
 L = 42.2 ohm.
The impedance of the circuit after inserting the capacitance is given by
 2  1  
2
Z=  R +  L −  
  C  
R R
Now, the power factor is given by cos  = =
Z  2  1  
2
R +  L −  
  C  
1 1
Clearly, for making power factor = 1.0, it must be that L = or C =
C (L)
But  = 2f = 2 × 3.14 × 50 = 314
1
C= = 75 × 10–6 farad
314  42.2
= 75 micro farad.

20. The image of object O from mirror M1 is I1 and the image of I1 (the virtual object) from mirror M2 is I3. The
image of object O from mirror M2 is I2 and the image of I2 (the virtual object) from mirror M1 is I4. Notice
that this interpretation, according to ray diagram rules, is valid only for Fig. (A). All others are inconsistent.
Hence correct is (A)

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21. 5 / 2(5 ) 5
= x = =
10 4

i.e  =
2
1
I = I 0C = I 0 / 2
2

22. 1

dU Vo
U = Vo ln (r), therefore, F = − =
dr r

If v is the orbital speed of the electron and m its mass, then

mv 2 Vo
=  mv 2 = Vo (1)
r r

Bohr’s quantum condition is mvr = nh/2

v = nh/2mr (2)

 h2  2
Using (2) in (1), we get r 2 =  n
 4 mVo 
2

r  n, Hence x = 1.

23. 2
The energy of released in transmission
1 1 7
4 → 3 is 32 13.6  2 − 2  = 13.6
 3 4  16
= 5.95eV
Work function = 5.95 − 3.95
= 2 eV

mD + m p 3
24. Minimum energy require = BE  = 2 106 
mD 2
= 310 eV .
6

25. 0

In one dimensional elastic collision of two bodies of equal masses, the initial velocities of
bodies are interchanged after collision.

Therefore, velocity of the neutron after collision is zero.

Hence, it has zero energy.

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