Basic Science Test 15

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Basic Science Test 15

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Question 1
1 / -0

On a $$60cm*60cm$$ carrom board, a carrom-coin placed at the centre is hit by the striker. The coin moves to the front edge, rebounds and strikes to the mid-point of the right side limb and stops. The displacement of the coin is

A
$$30\sqrt{2}cm$$

B
$$30+30\sqrt{2}cm$$

C
$$30cm$$

D
$$60cm$$

Solution

Displacement means shortest path,
So, $$AC$$ is shortest path here.
$$AC=\sqrt{(30cm)^2+(30cm)^2}$$
$$=\sqrt{900+900}$$
$$=\sqrt{1800}$$
$$=30\sqrt{2}cm$$
Question 2
1 / -0

A particle goes from $$(0, 0)$$ to $$(0,8)$$ and from $$(0,8)$$ to $$(6, 8)$$. Distance travelled and displacement respectively are

A
$$8$$ and $$14$$

B
$$10$$ and $$14$$

C
$$10$$ and $$8$$

D
$$14$$ and $$10$$
Question 3
1 / -0

If the period of oscillation of a simple pendulum is increased by $$a\%$$ then the percentage in its length is

A
$$a/2$$

B
$$a$$

C
$$2a$$

D
$$1/2$$
Question 4
1 / -0

A very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is 5 m then the number of ball thrown per minute is (g = 10 m s$$^{-2}$$)

A
120

B
80

C
60

D
40

Solution
Question 5
1 / -0

In the system shown below calculate the reading of the weighing machine, if the man pulls the string with a force of $$400$$ N. (Given the mass of man is $$30$$ kg and the mass of elevator is $$10$$ kg, weighing machine is light and string and pulley are ideal).

A
$$30\,kg$$

B
$$20\,kg$$

C
$$10\,kg$$

D
Zero
Question 6
1 / -0

A d-shell containing four unpaired electrons can exchange

A
four electrons

B
three electrons

C
sixteen electrons

D
six electrons

Solution
Question 7
1 / -0

Two bodies of masses $$5kg$$ and $$3kg$$ are moving towards each other with $$2ms^{-1}$$ and $$4ms^{-1}$$
respectively then velocity of centre of mass is

A
$$0.25 ms^{-1}\ towards \ 3\ kg$$

B
$$0.5 ms^{-1}\ towards \ 5 \ kg$$

C
$$0.25 ms^{-1}\ towards \ 5 \ kg$$

D
$$0.5 ms^{-1}\ towards \ 3 \ kg$$

Solution

We know that
$$\nu_{com}=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}$$
$$=\dfrac{(5)(+2)+(3)(-4)}{(5+3)}$$
$$=\dfrac{10-12}{8}$$
$$=2/8$$
$$=-0.25 m/s$$
so $$0.25 m/s$$ towards $$5\ kg$$
Question 8
1 / -0

A car moving at a consatnt speed of $$36\ kmph$$ moves north wards for $$20$$ minutes then due to west with the same speed fro $$8\dfrac{1}{3}$$ minutes. What is the average velocity of the car during this run in kmph

A
$$27.5$$

B
$$40.5$$

C
$$20.8$$

D
$$32.7$$

Solution

Starting point $$\rightarrow 0$$
$$OA$$ distance covered at speed $$636 kmph$$ for $$20$$ minutes
$$AB$$ distance convert at some speed $$36\ kmph$$ for $$8 1/3$$ minutes $$=\dfrac{25}{3}min$$
$$\therefore OA=36 \times \dfrac{20}{60}=12 km$$
$$AB=36 \times \dfrac{(25/3)}{60}=\dfrac{6}{10} \times \dfrac{25}{3}=5\ km$$
$$\therefore OB^2=OA^2+AB^2=12^2+5^2=169$$
$$\Rightarrow OB=\sqrt{169}=13\ km$$
$$\therefore$$ Displacement $$=OB=13 km$$
Total time $$= 20+\dfrac{25}{3}=\dfrac{85}{3}minutes$$
$$\therefore$$ Avg. velocity $$=\dfrac{Displacement}{total\ time}=\dfrac{13\ km}{\dfrac{(85/3)}{60} hr}=27.5 kmph$$
Question 9
1 / -0

A car travelling at a speed of 30 kilometer per hour is brought to a halt in 8 metres by applying brakes. If the same car is travelling at 60 km per hour, it can be brought to a half with same braking power in

A
$$8$$ metres

B
$$16$$ metres

C
$$24$$ metres

D
$$32$$ metres

Solution

Applying kinetic energy theorem
$$\Delta KE = \Delta w$$$$\frac{1}{2}mv^2 = F \cdot s$$
$$v^2 \alpha s$$
$$\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{s_1}{S_2}$$
$$v_2 = 2v_1$$
$$= \frac{1}{4} = \frac{s_1}{S_2}$$
$$s_2 = 4s_1$$
$$= 4 \times 8$$
$$= 32$$
Hence (A) option is correct answer
Question 10
1 / -0

A stone is dropped from the top of a tower of height h. After 1 s another stone is dropped from the balcony $$20\,m$$ below the top. Both reach the bottom simultaneously. What is the value of h ? Take $$g = 10\,ms^{-2}.$$

A
$$3125\,m$$

B
$$312.5\,m$$

C
$$31.25\,m$$

D
$$25.31\,m$$

Solution

Ball 1
$$-h = ut - \dfrac{1}{2}gt^2$$
$$\Rightarrow h = \dfrac{1}{2}gt^2$$ .....(i)
Ball 2
$$-(h - 20) = u(t - 1) - \dfrac{1}{2}g(t - 1)^2$$
$$\Rightarrow h - 20 = \dfrac{1}{2}g(t - 1)^2$$ ........(ii)
(i) & (ii) $$\Rightarrow h - 20 = \dfrac{1}{2} g \left ( \sqrt{\dfrac{2h}{g}} - 1 \right )^2$$
$$h - 20 = \dfrac{1}{2}g \left ( \dfrac{2h}{g} + 1 - 2 \times \sqrt{\dfrac{2h}{g}} \right )$$
$$h - 20 = h + \dfrac{g}{2} - \dfrac{g}{2} \times 2\dfrac{\sqrt{2h}}{\sqrt g}$$
$$\Rightarrow -20 = 5 - \sqrt{2hg}$$
$$\Rightarrow \sqrt{2hg} = 25 \Rightarrow 2hg = 625$$
$$\Rightarrow h = \dfrac{625}{20} = 31.25\,m$$