DSSSB PGT Physics Mock Test - 9 - DSSSB TGT/PGT/PRT MCQ

How the number is obtained?

2 + 4 = 6

5 + 9 = 14

Similarly,

3 + 5 = 8

Therefore, the answer is K₈.

Correct Answer :- B

Explanation : G×T+Q÷M

G(male) is father of T

T(female) is sister of Q

Q(female) is daughter of M

From the equations, we came to know, M is the wife of G.

V occurs at: 11 ;21 ;24 ;41 ;44 ;
A occurs at: 02 ;03 ;14 ;20 ;31 ;
R occurs at: 65 ;66 ;75 ;89 ;99 ;
Y occurs at: 57 ;58 ;67 ;76 ;97 ;

Sudhir's position from left = (12 + 4) = 16th from left
Clearly, there are three students between Ritesh and Sudir
∴ Total Number of children = (12 + 3 + 22) = 37

So Option C is correct

The figure may be labelled as shown.

The simplest triangles are BFG, CGH, EFM, FMG, GMN, GHN, HNI, LMK, MNK and KNJ i.e. 10 in number.

The triangles composed of three components each are FAK and HKD i.e. 2 in number.

The triangles composed of four components each are BEN, CMI, GLJ and FHK i.e. 4 in number.

The triangles composed of eight components each are BAJ and OLD i.e. 2 in number.

Thus, there are 10 + 2 + 4 + 2 = 18 triangles in the given figure.

From figures (i), (ii), (iv) we conclude that 5, 6, 1 and 2 lie adjacent to 4. Hence, 3 must lie opposite 4 and vice-versa. In fig. (iii), 3 is at the top and consequently 4 must lie at the bottom face.

The figure may be labelled as shown.

The simplest triangles are AHL, LHG, GHM, HMB, GMF, BMF, BIF, CIF, FNC, CNJ, FNE, NEJ, EKJ and JKD i.e. 14 in number.

The triangles composed of two components each are AGH, BHG, HBF, BFG, HFG, BCF, CJF, CJE, JEF, CFE and JED i.e. 11 in number.

The triangles composed of four components each are ABG, CBG, BCE and CED i.e. 4 in number.

Total number of triangles in the given figure = 14 + 11 + 4 = 29.

M occurs at: 13 ;22 ;33 ;42 ;43 ;
A occurs at: 01 ;03 ;11 ;32 ;40 ;
T occurs at: 55 ;65 ;67 ;68 ;78 ;
H occurs at: 85 ;87 ;95 ;96 ;98 ;

Parallactic second is the unit of distance because parallactic second is an abbreviation of parsec. Parsec = Parsec is the Unit for larger distances.It is the distance at which a star would make parallax of one Second of arc.

The newton second (also newton-second, symbol N s or N.s) is the derived SI unit of impulse. It is dimensionally equivalent to the momentum unit kilogram metre per second (kg.m/s). One newton second corresponds to a one-newton force applied for one second.

F = ma
[F] = kg m/s²

Unit of velocity = m/s
1m = 10⁶  Micro meter
And 1 sec = 10⁶  Micro sec
Thus 1 m/s = 10⁶ 10⁶  micro meter /micro second

Dimension of KE is ML²T^-2
That of pressure is ML^-2T^-2
That of momentum is ML/T 
That of power is ML²T^-3

Dimension of work and energy are the same i.e. [ML²T⁻²].
 Power = Energy/Time,
 Energy = Power × Time (in terms of hour instead of second),
 Energy = Watt × Hour,
= [ML²T⁻³] [T],
= [ML²T⁻²].
But, Momentum(p) = Mass × Velocity,
 p = [M] [LT⁻¹],
 p = [MLT⁻¹].
Hence the Dimension of work, energy and watt-hour is the same i.e. [ML²T⁻²] but the dimension of momentum is different ([MLT⁻¹]).

S_P​(Heat) = mcΔt

If radian correction is not considered then Δt will increase.

So, S_P​ will also increase.

Hence option 'A' is correct.

 Mass of Gold pieces = 2 × 0.035 = 0.070 g
 Mass of box = 2.3 g
 Total mass of box with gold pieces = 0.070 + 2.3 = 2.370 g

Since the answer should be in 2 significant digits, therefore, on rounding off, we get 2.4 g.

Thus, the required answer will be 2.4 g.

The number of significant figures is 3 because 10³ is a decimal multiplier.

So. after calculating this number, we will get 0310 which is also a 3 significant figure number. 

The astronomers mainly use parallax method to find distances to objects like other planets. To calculate the distance to a star or to a planet, astronomers observe it from different places along Earth's orbit around the Sun.

1 revolution in 40 seconds
In 1 second it covers = 1/40 revolution
In 140 sec = 1/40 * 140 =7/2 rotation = 3 and half rotation

Then the particle will be on the diametrically opposite end. 

Therefore, Displacement = R + R = 2R

• When an object is moving along a straight path, the magnitude of average velocity is equal to the average speed.
• Therefore, the numerical ratio of average velocity to average speed is one in a straight line motion.
• But, during curved motion, the displacement < distance covered, so the velocity < speed.

​Therefore, the numerical ratio of average velocity to average speed is equal to or less than one but can never be more than one. 

As the particles are moving at the same rate, their speed is the same but velocities will be different beacuse their directions will be different.

v = ds/dt

v = 12t - 3t²

0 = 3t (4 - t)
⇒ t = 0 or 4
Here t = 0 is not realistic so at t = 4 the velocity will be zero.
Thus, the particle will attain zero velocity at the time of 4 seconds.

Given:

Let two balls A and B have mass m_A, m_B respectively, and their initial velocities are u_A and u_B. After the collision, they will move with the same velocity, v_o. 

Given that mass of both balls are same.

So  m_A = m_B = m

u_A = V,  u_B = 0

From the Concept of Momentum Conservation:

m_Au_A + m_Bu_B = (m+m)v_o

mV = 2mv_o

v_o = V/2

Both the balls after impact will move with velocity v/2.