DSSSB PGT Physics Mock Test - 8 - DSSSB TGT/PGT/PRT MCQ

(7 x 3) = 21 and (9 x 3) = 27

and (4 x 9) = 36 and (2 x 9) = 18

Therefore (9 x 6) = 54 and (4 x 6) = 24.

(16 + 13) = (14 + 15)

and (28 + 12) = (10 + 30)

Therefore (29 + 16) = (15 + 30).

(3 x 4) + 3 = 15

and (7 x 5) + 3 = 38

Therefore (3 x 5) + 3 = 18.

(15 - 12) + (10 - 9) = 4

(28 - 12) + (16 - 20) = 12

Similarly, (23 -11) + (15- 16) = 11.

As 2012 was a leap year, so the number of days in each month from 23rd January, 2011 to 31st July, 2013 is given as follows.
Since 2011 is not a leap year = 365 - 22 = 343 days
Number of days in 2012 = 366 days (leap year)
Number of days in 2013:
January = 31 days
February = 28 days
March = 31 days
April = 30 days
May = 31 days
June = 30 days
July = 31 days
Total number of days in 2013 = 212
Therefore, total number of days from 23rd January, 2011 to 31st July, 2013 = 343 + 366 + 212 = 921

The time duration from 10 a.m. on Friday to 2 p.m. on the following Monday is 76 hours.
The clock gains 1 minute per hour.
∴ Time gained in 76 hours = 76 minutes = 1 hour 16 minutes
∴ Time shown by the second clock will be 3:16 p.m.

He returned after 8 o'clock but before 9 o'clock, at such a time when both the minute hand and the hour hand were found to be coinciding.
In a simple way, we need to calculate the time when the angle between the minute hand and the hour hand is zero.
Angle between the minute hand and the hour hand at 8 o'clock = 8 × 30 = 240°
We can reduce the difference of 5.5 degree in 1 min,
Required time = 8 hrs + 240 × 2 ÷ 11

In 1 hour, the first clock turns 62 minutes and the second 56 minutes. This implies that for every 56 minutes of the second clock, the first clock gains 6 minutes.
'The second clock shows 10 p.m.' means that it runs for 14 hours.
In this time, the first clock gains = 90 minutes.
So, the time will be 11:30 p.m.

To be together between 9 and 10 o'clock, the minute hand has to gain 45 minute spaces.
55 minute space is gained in 60 minutes.
45 minute space will be gained in 
At 49minutes past 9, the hands of the clock will be together.

According to given information:

Statements: M ≤ N = O ; M ≤ P ; O < Q

Combining the statement: P > M ≤ N = O < Q

Here, P is greater than or equal to M which is less than or equal to N equal to O which is less than Q.

Conclusions:

I.  Q > N → True (In the given statement, P > M ≤ N = O < Q ,It clearly mentioned Q is greater than O which is equal to N .So, Q is greater than N too. Thus, the conclusion I is true).

II. M ≥ O → False (In the given statement, M ≤ N = O, It is clearly mentioned that M is less than or equal to N which is equal to O. So, M can never be greater than O. Thus, conclusion II is a false).

Hence, "Option b" is the correct answer.

The given statement is: J = K ≥ L > X ≥ Y ;  J ≥ P ; P ≥ Z 

Combining the statement: Z < P < J = K ≥ L > X ≥ Y 

Here, Z is smaller than or equal to P which is Smaller than or equal to J which is equal to K which is greater than or equal to L which is greater than X and X is greater than or equal to Y.

Conclusions:

I: J < L → False (as J = K ≥ L gives as J = K and K ≥ L which implies that J ≥ L and J cannot be less than L . Thus, Conclusion I is false.).

II:  P ≤ Y → False (as Z < P < J = K ≥ L > X ≥ Y  → clear relationship between P and Y cannot be determined. Thus, Conclusion II is false.)

Hence, the "Option c" is the correct answer.

According to given information:

Statements: S = R ≥ Q ≤ P ; N ≤ P ; M < N

Combine the statement: S = R ≥ Q ≤ P  ≥ N > M

Here, S is equal to R which is greater than or equal to Q which is less than or equal to P which is greater than or equal to N which is greater than M.

Conclusions:

I. M < S → False (As S = R ≥ Q ≤ P ; N ≤ P ; M < N → there is no direct relation between M and S. So, conclusion I is false)

II. M ≥ S → False (As S = R ≥ Q ≤ P ; N ≤ P ; M < N → there is no direct relation between M and S. So, conclusion II is false)

Here, both conclusions I and II form a complementary pair as both have the same element i.e. M and S and all the three symbols (<, =, >) are present between M and S.

Hence, Either I or II follows.

Hence, "Option d" is the correct answer.

Given: A > B, B > C and C > D

After combining the operators,

we get: A > B > C > D

Now, checking the options:

Option A: A > D → True (As, A > B > C > D gives A > D)

Option B: A > C → True (As, A > B > C > D gives A > C)

Option C: D > A → False (As, A > B > C > D gives A > D)

Option D: B > D → True (As, A > B > C > D gives B > D)

So, Option C, D > A is definitely wrong.

Hence, the correct answer is "Option c".

As the result of Race is Fatigue similarly the result of Fast is Hunger.

Dimensional Analysis (also called Factor-Label Method or the Unit Factor Method) is a problem-solving method that uses the fact that any number or expression can be multiplied by one without changing its value. It is a useful technique.

Another use of dimensional analysis is in checking the correctness of an equation which you have derived after some algebraic manipulation. Even a minor error in algebra can be detected because it will often result in an equation which is dimensionally incorrect.

The rules for rounding off are following:

• If the first non-significant digit is less than 5, then the least significant digit remains unchanged.
• If the first non-significant digit is greater than 5, the least significant digit is incremented by 1.
• If the first non-significant digit is 5, the least significant digit can either be incremented or left unchanged.
• All non-significant digits are removed.

So rounding off 34.216 upto 4 digits is 34.22 and upto 3 digits is 34.2

Resolution is the smallest amount of input signal change that the instrument can detect reliably. This term is determined by the instrument noise (either circuit or quantization noise).

If an equation fails this consistency test, it is proved wrong, but if it passes, it is not proved right.

Thus, a dimensionally correct equation need not be actually an exact (correct) equation, but a dimensionally wrong (incorrect) or inconsistent equation is definitely wrong.

We define length and time separately as it is not possible to define velocity without using these quantities. This means that one fundamental quantity depends on the other. So, these quantities cannot be listed as fundamental quantities in any system of units.

v²r/g= (L¹T^-1)2L1/LT^-2=L²
v²/rg= (LT^-1)² L1(L¹ T^-2) ​=M⁰L⁰T⁰
v²g/r​=(LT^-1)2 LT^-2/L​=L²T^-4
v²rg=(LT^-1)²(L1) (LT^-1) =L⁴T^-3
So, option D is correct.

G = 6.67 x 10-11 Nm2/kg2

so, when expressed in VGS units we shall convert N to dynes, m to cm and kg to g, thus

G = 6.67 x 10-11 x [(105dynes x (102)2cm2 ) / (103)2 g2]

=  6.67 x 10-11 x [ (105 x 104 ) / 106 ] 

Thus,

in CGS system

G = 6.67 x 10-8 dyne.cm2/g2

• Dimensional Equations are used to derive the Relations between the Physical quantities.
• Also, they are used to check the correctness of the Equation by the Principle of the homogeneity of the dimensions.

Hence, Option (a) and option (b) are true for the Dimensional equation. Thus, It is not correct for the given questions.

• Dimensional Equations are not valid for Pure numbers since they are non-dimensional constants. Also, they do not provide the value of the Constant of the Proportionality.
• Thus, we can also say that all equations which are correct must be dimensionally correct but all those equations that are dimensionally correct may or may not be accurately correct. Thus, Option (c) is the correct answer for the given question.
• Dimensional Equations are used for the conversions of the units thus (d) option can not be correct for the given question.

Hence, option C is the correct answer as it is an incorrect statement. 

The significant figures in any number are all certain digits plus one doubtful digit. Certain rules are followed which are as under:

1) All one zero digits in a number are significant.

2) The zeroes between two non-zero digits are always significant.

3) The zeros written to the left of the first non-zero digit in a number are non-significant. They simply indicate the position of the decimal point.

4) All zeros placed to the right of a decimal point in a number are significant.

Since, 0.0268 has three significant figures whereas 2.608 cm, 26.08 mm and 2.068 cm have four significant figures, A is the correct answer. 

Cesium 133 is the element most commonly chosen for atomic clocks.

Another solution :-

• Displacement is the shortest length between the starting and the end pointing of a journey, whereas the distance is the actual length of the path travelled.
• Hence, the distance will always be equal to or greater than the displacement.
• So, the ratio of displacement to distance will be always less than or equal to 1, since the denominator(distance) is either greater than or equal to the numerator (displacement).

Let the vehicle travels from A to B. Distances, velocities, and time taken are shown. To calculate average speed we will calculate the total distance covered and will divide it by the time interval in which it covers that total distance. 

Time taken to travel first half distance t₁
Time taken to travel first half distance t₁

Total time taken = t₁ + t₂

We know that v_av ​= Average speed = Total Distance / Total Time

The correct option is B Both cannot be zero
If the location of a particle has changed then it certainly did some displacement and if it has undergone some displacement, then it has travelled some distance too.

This can be verified graphically

ObtaIning instantaneous acceleration from graph