Basic Science Test 16

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Basic Science Test 16

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Weekly Quiz Competition

Question 1
1 / -0

A particle is projected with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is acceleraction due to gravity, then its range is:

A
$$\dfrac{4v^2}{5g}$$

B
$$\dfrac{4g}{5v^2}$$

C
$$\dfrac{4v^3}{5g^2}$$

D
$$\dfrac{4v}{5g^2}$$

Solution
Question 2
1 / -0

A heavy box with mass M 200 kg slides down a 4 m long ramp which makes $$30^o$$ with horizontal. There is a small friction force which performs work W-2250 J. The box had no initial velocity at the top of the ramp. Find the speed of the box (in m/s) at the bottom of the ramp.

A
1

B
2

C
3

D
4

E
5

Solution

Initially body is at rest. So it posses only P.E
Since total energy has to be conserved.
Loss in P.E = Gain in K.E + work done against friction
$$mgh - 2250 = \frac{1}{2} mv^2$$
$$(200)(10) (4 \sin 30^o) - 2250 = 100v^2$$
$$(200)(2) - 2250 = 100v^2$$
$$4000 - 2250 = 100v^2$$
$$1750 = 100 v^2$$
$$\sqrt{17.5} = v$$
$$v = 4m/s$$
Question 3
1 / -0

A body slides down a smooth inclined plane of base $$x$$ meter. Find the angle of the plane so that the time taken by the body to reach the bottom of the plane is minimum

A
$$60^{\circ}$$

B
$$30^{\circ}$$

C
$$45^{\circ}$$

D
$$53^{\circ}$$

Solution

$$bass \rightarrow x meter$$
$$cos \theta = \dfrac{x}{d}$$
$$d = (\dfrac{x}{cos\theta })$$
$$d \rightarrow $$$$S = ut + \dfrac{1}{2}a t^2$$$$d = 0 + \dfrac{1}{2}g sin\theta t^2$$
$$\sqrt{\dfrac{2d}{gsin\theta } = t}$$
$$t \rightarrow $$ time taken $$\dfrac{dt}{d\theta } = 0$$
$$\dfrac{dt}{d\theta } = \sqrt{\dfrac{2d}{9}} \times \dfrac{1}{\sqrt{sin \theta }}$$
$$= (sin\theta )^{-1/2}$$
$$= -\dfrac{1}{2} (sin \theta )^{-3/2} \times cos \theta = 0$$
$$\dfrac{dx}{d\theta } = 0$$
$$\theta= 45$$
Question 4
1 / -0

A boat which can move with a speed of $$5\ m/s$$ relative to water crosses a river of width $$480\ m$$ flowing with a constant speed of $$4\ m/s$$. Find the time taken by the boat to cross the river along the path which is shortest. Find also the length of shortest path.

A
$$480\ s, 160\ m$$

B
$$160\ s, 480\ m$$

C
$$480\ s, 480\ m$$

D
$$160\ s, 160\ m$$

Solution

time taken by boat to cross along shortest paths
$$t=\dfrac {d}{\sqrt{V^2-U^2}}$$
$$d\to $$ width of river
$$v\to $$ speed of boat relative to river
$$v\to $$ speed of river
$$U=4\ m/s\ V=5\ m/s$$
$$d=480\ m$$
$$t=\dfrac {480}{\sqrt{5^2-4^2}}=160\ sec$$
Question 5
1 / -0

A 100 m long train crosses a man travelling at 5 km/h, in opposite direction in 7.2 seconds, then the velocity of train is:-

A
40 km/h.

B
25 km/h.

C
20 km/h.

D
45 km/h.

Solution

$$u = 5 km/h = 5\times\dfrac{5}{18} = \dfrac{25}{18} m/s.$$
Velocity = $$\dfrac{distance}{time}$$
$$v + \dfrac{25}{18} = \dfrac{100}{7.2}$$
$$v + 1.388 = 13.88$$
$$v = 12.49 m/s$$
$$v = 12.49\times\dfrac{18}{5} = 45 km/h.$$
Question 6
1 / -0

A cyclist goes round a circular path of circumference $$\pi \ m$$ in $$\sqrt 2\ s$$, the angle made by him with the vertical will be

A
$$45^o$$

B
$$30^o$$

C
$$60^o$$

D
$$37^o$$

Solution

$$c=\pi m$$
$$r=\dfrac {\pi}{2\pi}=\dfrac 12m$$
$$T=\sqrt 2\ s$$
$$\omega =\dfrac {2\pi}{T}=\dfrac {2\pi}{\sqrt 2}\Rightarrow \sqrt 2\pi$$
$$\theta =\tan^{-1}\left(\dfrac {r\omega^2}{g}\right)$$
$$=\tan^{-1}\left( \dfrac {1\times 2\times \pi^2}{2\times 9.8}\right)$$
$$=\tan^{-1}\left( \dfrac {9.8}{9.8}\right)$$
$$=\tan^{-1}(1)$$
$$=45^o$$
Question 7
1 / -0

A physical quantity Q is found to depend on observable x, y and z obeying relation $$ Q= \frac {x^3 y^2}{z} $$ the percentage error in the measurements of x, y and z are $$ 1$$ %, $$ 2$$ % and $$ 4$$ % respectively. what is percentage error is the quantity Q.

A
$$ 11 $$ %

B
$$ 4 $$ %

C
$$ 1 $$ %

D
$$ 3 $$ %

Solution

$$ Q = \dfrac {x^3 y^2}{ z}$$
$$ \dfrac { \Delta Q}{Q} \times 100 = 3 \dfrac { \Delta x}x \times 100 + 3 \times \dfrac { \Delta y}{y} \times 100 + \dfrac { \Delta Z}{z} \times 100 $$

$$ \Delta Q= 3 \times (1) +2 (2) + \Delta $$

$$ \Delta Q= 3+ 4 + 4 = 11 $$ %
Question 8
1 / -0

In a vernier callipers , one scale division in x cm , and n division of the vernier scale coincide with (n -1) divisions of the main scale . The least count (in cm ) of the callipers is

A
$$ \left ( \dfrac{n -1}{n} \right )x $$

B
$$ \dfrac{nx}{(n -1)} $$

C
$$ \dfrac{x}{n} $$

D
$$ \dfrac{x}{(n- 1)} $$

Solution
Question 9
1 / -0

Maximum percentage error in the measurement of 25.00 m is

A
0.01%

B
0.03%

C
0.04%

D
0.02%

Solution

L = 25.00 meter
maximum error = 1 cm
$$ \dfrac{1cm}{25m}\times{100} $$
= $$ {10}_{-2}\times{4} % $$
= 0.04%
Question 10
1 / -0

The dimension of heat capacity are

A
$$ \left [ ML^{-2} T^{-2}K^{-1} \right ] $$

B
$$ \left [ ML^{2} T^{-2}K^{-1} \right ] $$

C
$$ \left [ M^{-1}L^{2} T^{-2}K^{-1} \right ] $$

D
$$\left [ MLT^{-2} K \right ] $$

Solution

Answer
(Heat capacity) = $$ = \left [ \dfrac{Energy}{Temeperature} \right ] $$
$$ = \left [ \dfrac{F \times l}{k} \right ] = \dfrac{MLT^{-2} \times L}{K} $$ $$ML^{2}T^{-2}K^{-1} $$

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Basic Science Test 16

Result

Basic Science Test 16

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Question 1

$$\dfrac{4v^2}{5g}$$

$$\dfrac{4g}{5v^2}$$

$$\dfrac{4v^3}{5g^2}$$

$$\dfrac{4v}{5g^2}$$

Solution

Question 2

1

2

3

4

5

Solution

Question 3

$$60^{\circ}$$

$$30^{\circ}$$

$$45^{\circ}$$

$$53^{\circ}$$

Solution

Question 4

$$480\ s, 160\ m$$

$$160\ s, 480\ m$$

$$480\ s, 480\ m$$

$$160\ s, 160\ m$$

Solution

Question 5

40 km/h.

25 km/h.

20 km/h.

45 km/h.

Solution

Question 6

$$45^o$$

$$30^o$$

$$60^o$$

$$37^o$$

Solution

Question 7

$$ 11 $$ %

$$ 4 $$ %

$$ 1 $$ %

$$ 3 $$ %

Solution

Question 8

$$ \left ( \dfrac{n -1}{n} \right )x $$

$$ \dfrac{nx}{(n -1)} $$

$$ \dfrac{x}{n} $$

$$ \dfrac{x}{(n- 1)} $$

Solution

Question 9

0.01%

0.03%

0.04%

0.02%

Solution

Question 10

$$ \left [ ML^{-2} T^{-2}K^{-1} \right ] $$

$$ \left [ ML^{2} T^{-2}K^{-1} \right ] $$

$$ \left [ M^{-1}L^{2} T^{-2}K^{-1} \right ] $$

$$\left [ MLT^{-2} K \right ] $$

Solution

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