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TEST SERIES (PHYSICS FST)

266596 In the alternating circuit shown in figure the current measured by ammeter is :

1 1 A

2 2 A

3 3 A

4 zero

Explanation:

b
$\begin{aligned} 50 = \sqrt{(100 - 60)^{2} + V_{R^{2}}} \\ 2500 = 1600 + V_{R^{2}} \\ V_{R^{2}} = 900 \\ V_{R} = 30 V \\ i = \frac{30}{15} = 2 A \end{aligned}$

TEST SERIES (PHYSICS FST)

266597 A body is projected vertically in upward direction from ground with speed $20 m / s$ . It will come on ground after $(g = 10 m / s^{2})$ :

1 2 sec

2 4 sec

3 20 sec

4 12 sec

Explanation:

b
$\begin{aligned} v = u - g t \\ 0 = 20 - 10 t \\ t = 2 \sec \end{aligned}$
So total time of flight $= 2 t$
$= 2 \times 2 = 4 \sec .$

TEST SERIES (PHYSICS FST)

266598 A body $A$ starts from rest with an accleration $a_{1}$ . After 2 seconds, another body B starts from rest with an acceleration $a_{2}$ . If they travel equal distance in the 5 th second, after the start of $A$ , then the ratio $a_{1} : a_{2}$ is equal to:

1

5 : 9

2

5 : 7

3

9 : 5

4

9 : 7

Explanation:

a
Time taken by body $A_{1} t_{1} = 5 \sec$ .
Time taken by body B, $t_{2} = 5 - 2 = 3 \sec$ .
Distance covered by first body in 5th second after start.
$s_{5} = u + \frac{a_{1}}{2} {2 t_{1} - 1} \Rightarrow 0 + \frac{a_{1}}{2} {2 \times 5 - 1} = \frac{5 a_{1}}{2}$
$s_{3} = u + \frac{a_{2}}{2} {2 t_{2} - 1} \Rightarrow 0 + \frac{a_{2}}{2} {2 \times 3 - 1} = \frac{5 a_{2}}{2}$
Since, $s_{5} = s_{3}$
$∴ \frac{9 a_{1}}{2} = \frac{5 a_{2}}{2}$ or $a_{1} : a_{2} = 5 : 9$ .

TEST SERIES (PHYSICS FST)

266599 A bullet is fired horizontally with a speed $500 m / s$ in order to hit a target 50 m away. If $g = 10 m / s^{2}$ the gun should be aimed :

1 Directly towards the target

2 5 cm above the target

3 10 cm above the target

4 15 cm above the target

Explanation:

b

vertical drift $y = \frac{1}{2} g t^{2} - - (1)$
Horizontal displacement $x = u \times t \Rightarrow t = \frac{x}{u}$
fromeq. (1)
$\begin{aligned} y = \frac{1}{2} g {(\frac{x}{u})}^{2} \\ y = \frac{1}{2} \times 10 \times \frac{2500}{250000} \\ y = \frac{5}{100} m \\ \Rightarrow y = \frac{5}{100} \times 100 cm \\ = y = 5 cm \end{aligned}$

TEST SERIES (PHYSICS FST)

266596 In the alternating circuit shown in figure the current measured by ammeter is :

1 1 A

2 2 A

3 3 A

4 zero

Explanation:

b
$\begin{aligned} 50 = \sqrt{(100 - 60)^{2} + V_{R^{2}}} \\ 2500 = 1600 + V_{R^{2}} \\ V_{R^{2}} = 900 \\ V_{R} = 30 V \\ i = \frac{30}{15} = 2 A \end{aligned}$

TEST SERIES (PHYSICS FST)

266597 A body is projected vertically in upward direction from ground with speed $20 m / s$ . It will come on ground after $(g = 10 m / s^{2})$ :

1 2 sec

2 4 sec

3 20 sec

4 12 sec

Explanation:

b
$\begin{aligned} v = u - g t \\ 0 = 20 - 10 t \\ t = 2 \sec \end{aligned}$
So total time of flight $= 2 t$
$= 2 \times 2 = 4 \sec .$

TEST SERIES (PHYSICS FST)

266598 A body $A$ starts from rest with an accleration $a_{1}$ . After 2 seconds, another body B starts from rest with an acceleration $a_{2}$ . If they travel equal distance in the 5 th second, after the start of $A$ , then the ratio $a_{1} : a_{2}$ is equal to:

1

5 : 9

2

5 : 7

3

9 : 5

4

9 : 7

Explanation:

a
Time taken by body $A_{1} t_{1} = 5 \sec$ .
Time taken by body B, $t_{2} = 5 - 2 = 3 \sec$ .
Distance covered by first body in 5th second after start.
$s_{5} = u + \frac{a_{1}}{2} {2 t_{1} - 1} \Rightarrow 0 + \frac{a_{1}}{2} {2 \times 5 - 1} = \frac{5 a_{1}}{2}$
$s_{3} = u + \frac{a_{2}}{2} {2 t_{2} - 1} \Rightarrow 0 + \frac{a_{2}}{2} {2 \times 3 - 1} = \frac{5 a_{2}}{2}$
Since, $s_{5} = s_{3}$
$∴ \frac{9 a_{1}}{2} = \frac{5 a_{2}}{2}$ or $a_{1} : a_{2} = 5 : 9$ .

TEST SERIES (PHYSICS FST)

266599 A bullet is fired horizontally with a speed $500 m / s$ in order to hit a target 50 m away. If $g = 10 m / s^{2}$ the gun should be aimed :

1 Directly towards the target

2 5 cm above the target

3 10 cm above the target

4 15 cm above the target

Explanation:

b

vertical drift $y = \frac{1}{2} g t^{2} - - (1)$
Horizontal displacement $x = u \times t \Rightarrow t = \frac{x}{u}$
fromeq. (1)
$\begin{aligned} y = \frac{1}{2} g {(\frac{x}{u})}^{2} \\ y = \frac{1}{2} \times 10 \times \frac{2500}{250000} \\ y = \frac{5}{100} m \\ \Rightarrow y = \frac{5}{100} \times 100 cm \\ = y = 5 cm \end{aligned}$

TEST SERIES (PHYSICS FST)

266596 In the alternating circuit shown in figure the current measured by ammeter is :

1 1 A

2 2 A

3 3 A

4 zero

Explanation:

b
$\begin{aligned} 50 = \sqrt{(100 - 60)^{2} + V_{R^{2}}} \\ 2500 = 1600 + V_{R^{2}} \\ V_{R^{2}} = 900 \\ V_{R} = 30 V \\ i = \frac{30}{15} = 2 A \end{aligned}$

TEST SERIES (PHYSICS FST)

266597 A body is projected vertically in upward direction from ground with speed $20 m / s$ . It will come on ground after $(g = 10 m / s^{2})$ :

1 2 sec

2 4 sec

3 20 sec

4 12 sec

Explanation:

b
$\begin{aligned} v = u - g t \\ 0 = 20 - 10 t \\ t = 2 \sec \end{aligned}$
So total time of flight $= 2 t$
$= 2 \times 2 = 4 \sec .$

TEST SERIES (PHYSICS FST)

266598 A body $A$ starts from rest with an accleration $a_{1}$ . After 2 seconds, another body B starts from rest with an acceleration $a_{2}$ . If they travel equal distance in the 5 th second, after the start of $A$ , then the ratio $a_{1} : a_{2}$ is equal to:

1

5 : 9

2

5 : 7

3

9 : 5

4

9 : 7

Explanation:

a
Time taken by body $A_{1} t_{1} = 5 \sec$ .
Time taken by body B, $t_{2} = 5 - 2 = 3 \sec$ .
Distance covered by first body in 5th second after start.
$s_{5} = u + \frac{a_{1}}{2} {2 t_{1} - 1} \Rightarrow 0 + \frac{a_{1}}{2} {2 \times 5 - 1} = \frac{5 a_{1}}{2}$
$s_{3} = u + \frac{a_{2}}{2} {2 t_{2} - 1} \Rightarrow 0 + \frac{a_{2}}{2} {2 \times 3 - 1} = \frac{5 a_{2}}{2}$
Since, $s_{5} = s_{3}$
$∴ \frac{9 a_{1}}{2} = \frac{5 a_{2}}{2}$ or $a_{1} : a_{2} = 5 : 9$ .

TEST SERIES (PHYSICS FST)

266599 A bullet is fired horizontally with a speed $500 m / s$ in order to hit a target 50 m away. If $g = 10 m / s^{2}$ the gun should be aimed :

1 Directly towards the target

2 5 cm above the target

3 10 cm above the target

4 15 cm above the target

Explanation:

b

vertical drift $y = \frac{1}{2} g t^{2} - - (1)$
Horizontal displacement $x = u \times t \Rightarrow t = \frac{x}{u}$
fromeq. (1)
$\begin{aligned} y = \frac{1}{2} g {(\frac{x}{u})}^{2} \\ y = \frac{1}{2} \times 10 \times \frac{2500}{250000} \\ y = \frac{5}{100} m \\ \Rightarrow y = \frac{5}{100} \times 100 cm \\ = y = 5 cm \end{aligned}$

TEST SERIES (PHYSICS FST)

266596 In the alternating circuit shown in figure the current measured by ammeter is :

1 1 A

2 2 A

3 3 A

4 zero

Explanation:

b
$\begin{aligned} 50 = \sqrt{(100 - 60)^{2} + V_{R^{2}}} \\ 2500 = 1600 + V_{R^{2}} \\ V_{R^{2}} = 900 \\ V_{R} = 30 V \\ i = \frac{30}{15} = 2 A \end{aligned}$

TEST SERIES (PHYSICS FST)

266597 A body is projected vertically in upward direction from ground with speed $20 m / s$ . It will come on ground after $(g = 10 m / s^{2})$ :

1 2 sec

2 4 sec

3 20 sec

4 12 sec

Explanation:

b
$\begin{aligned} v = u - g t \\ 0 = 20 - 10 t \\ t = 2 \sec \end{aligned}$
So total time of flight $= 2 t$
$= 2 \times 2 = 4 \sec .$

TEST SERIES (PHYSICS FST)

266598 A body $A$ starts from rest with an accleration $a_{1}$ . After 2 seconds, another body B starts from rest with an acceleration $a_{2}$ . If they travel equal distance in the 5 th second, after the start of $A$ , then the ratio $a_{1} : a_{2}$ is equal to:

1

5 : 9

2

5 : 7

3

9 : 5

4

9 : 7

Explanation:

a
Time taken by body $A_{1} t_{1} = 5 \sec$ .
Time taken by body B, $t_{2} = 5 - 2 = 3 \sec$ .
Distance covered by first body in 5th second after start.
$s_{5} = u + \frac{a_{1}}{2} {2 t_{1} - 1} \Rightarrow 0 + \frac{a_{1}}{2} {2 \times 5 - 1} = \frac{5 a_{1}}{2}$
$s_{3} = u + \frac{a_{2}}{2} {2 t_{2} - 1} \Rightarrow 0 + \frac{a_{2}}{2} {2 \times 3 - 1} = \frac{5 a_{2}}{2}$
Since, $s_{5} = s_{3}$
$∴ \frac{9 a_{1}}{2} = \frac{5 a_{2}}{2}$ or $a_{1} : a_{2} = 5 : 9$ .

TEST SERIES (PHYSICS FST)

266599 A bullet is fired horizontally with a speed $500 m / s$ in order to hit a target 50 m away. If $g = 10 m / s^{2}$ the gun should be aimed :

1 Directly towards the target

2 5 cm above the target

3 10 cm above the target

4 15 cm above the target

Explanation:

b

vertical drift $y = \frac{1}{2} g t^{2} - - (1)$
Horizontal displacement $x = u \times t \Rightarrow t = \frac{x}{u}$
fromeq. (1)
$\begin{aligned} y = \frac{1}{2} g {(\frac{x}{u})}^{2} \\ y = \frac{1}{2} \times 10 \times \frac{2500}{250000} \\ y = \frac{5}{100} m \\ \Rightarrow y = \frac{5}{100} \times 100 cm \\ = y = 5 cm \end{aligned}$

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