Basic Science Test 26

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Basic Science Test 26

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Weekly Quiz Competition

Question 1
1 / -0

An emergency vehicle travels $$10$$ miles at a speed of $$50$$ miles per hour. How fast must the vehicle travel on the return trip if the round-trip travel time is to be $$20$$ minutes?

A
$$72$$ miles per hour

B
$$75$$ miles per hour

C
$$65$$ miles per hour

D
$$78$$ miles per hour

Solution

Speed = Distance ÷ Time
Distance given = 10 miles
Speed given = 50 miles per hour
Time = x
Find out time
50=10÷x
50x=10
x=1÷5 hours
x=12 minutes
Total trip = 20 minutes
remaining time = 8 minutes
Now,
Time = 8 min (8÷60) hours = (2÷15) hours
Speed = x
Distance = 10 miles (as we will just have to ride back)
Speed = Distance ÷ Time
x=(10)÷(2÷15)
x=(10)×(15÷2)
x=150÷2
x=75
The speed will have to be 75 miles per hour
Question 2
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An on jest is dropped in an unknown planet from height $$50\ m $$ it reaches the ground in $$2s$$ the acceleration due to gravity in this unknown planet is

A
$$g = 20\ ms^{-2} $$

B
$$g = 25\ ms^{-2} $$

C
$$g = 15\ ms^{-2}$$

D
$$g = 30\ ms^{-2} $$

Solution

The answer is b
g=25 ms−2
Explanation :
Acceleration due to gravity = height / time
Question 3
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A boy aims a target at a horizontal distance of $$60\ m $$. If the muzzle speed of bullet is $$600\ m/s $$, then the height above the target which he should aim is

A
$$5\ cm $$

B
$$10\ cm $$

C
$$15\ cm $$

D
$$20\ cm $$

Solution

$$D = s \times t $$
$$t = \dfrac{D}{S} = \dfrac{60\ m}{600} = 0.1 $$ sec
$$0.1 $$ sec is the time for bullet to travel.
In vertically,
$$S = ut + \dfrac{1}{2} at^{2} $$
$$h = 0 \times t + \dfrac{1}{2} \times 10 \times 0.1 \times 0.1 $$
$$ = \dfrac{5}{100} = 0.05\ m $$
$$h = 5\ cm $$
Question 4
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The angular velocity of the body changes from $$ \omega_{1} $$ to $$ \omega_{2} $$ without applying torque but by changing moment of inertia. The ratio of initial radius of gyration to the final radius of gyration is :

A
$$1/ \omega_{2} : 1/\omega_{1} $$

B
$$\omega^{2}_{2} : \omega^{2}_{1} $$

C
$$ \sqrt{\omega_{2}} : \sqrt{\omega_{1}} $$

D
$$ \omega_{2} : \omega_{1} $$

Solution

Radius of gyration , $$k = \
sqrt{\dfrac{I}{m}} $$

No change in torque
So angular momentum conserved
$$I_{1} w_{1} = I_{2} w_{2} $$
$$ \dfrac{I_1}{I_2} = \dfrac{w_2}{w_1} $$

$$ \sqrt{\dfrac{I_1}{I_2}} = \sqrt{\dfrac{w_2}{w_1}} $$

$$ \dfrac{k_1}{k_2} = \sqrt{\dfrac{I_1}{I_2}} = \sqrt{\dfrac{w_2}{w_1}} $$
Question 5
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A body is projected vertically upwards. The times corresponding to height $$H$$ while ascending and while descending are $$t_{1} $$ and $$t_{2} $$ respectively. Then the maximum height reached from the point of projection, is

A
$$ \dfrac{g(t_{1} + t_{2})^{2}}{4} $$

B
$$ \dfrac{g(t_{1} + t_{2})^{2}}{8} $$

C
$$ \dfrac{3}{8} g (t_{1} + t_{2})^{2} $$

D
$$ \dfrac{g (t_{1} + t_{2})^{2}}{2} $$

Solution

$$H = ut - \dfrac{1}{2} at^{2} $$
This has two roots
$$t_{1} $$ & $$t_{2} $$

$$H = ut_{1} - \dfrac{1}{2} at_{1}^{2} \Rightarrow \dfrac{H}{t_{1}} + \dfrac{1}{2} g \dfrac{t_{1}^{2}}{t_{1}} = 4 $$

$$H = ut_{2} - \dfrac{1}{2} at_{2}^{2} $$ $$ \Rightarrow \dfrac{H}{t_{2}} + \dfrac{1}{2} g \dfrac{t_{2}^{2}}{t_{2}} = 4 $$

$$u (t_{2} - t_{1}) = \dfrac{1}{2} g (t_{2}^{2} - t_{1}^{2}) $$

$$u = \dfrac{1}{2} g (t_{1} + t_{2}) $$

Max height = $$ \dfrac{u^2}{2g} = \dfrac{1}{4} g^{2} \dfrac{(t_{1} + t_{2})^{2}}{2g} $$

$$ = \dfrac{1}{8} g (t_{1} + t_{2})^{2} $$
Question 6
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A physical quantity $$X$$ is given by $$X=\dfrac{a^{3} b^{2} d}{c^{1 / 2}},$$ the percentage error in the measurement of $$a, b, c$$ and dare $$1 \%, 3 \%, 2 \%$$ and $$4 \%$$ respectively. The maximum percentage error in the measurement of $$X$$ is

A
$$10\%$$

B
$$14\%$$

C
$$17\%$$

D
$$19\%$$

Solution

$$\begin{array}{r} \% \dfrac{\Delta x}{x}=\left(3 \% \dfrac{\Delta a}{a}\right)+\left(2 \% \dfrac{\Delta b}{b}\right)+\left(\% \dfrac{\Delta d}{d}\right) +\left(\dfrac{1}{2} \% \dfrac{\Delta c}{c}\right) \end{array}$$

Given that

$$\%\dfrac{\Delta a}{a}=1\,\,\,\,\,\,\%\dfrac{\Delta b}{b}=2$$

$$\%\dfrac{\Delta c}{c}=3\,\,\,\,\,\,\, \%\dfrac{\Delta d}{d}=4$$

$$\dfrac{\Delta x}{x}=3 \times 1+2 \times 3+4+\dfrac{1}{2} \times 2$$

$$\dfrac{\Delta x}{x}=14\%$$
Question 7
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The side of a cube mesasured by a vernier calliper
whose 10 division of the veriner scale are equal to 8 divisions of main scale If the main scale reads 10 mm and second division of vernier scale coincides with one of the main scale division then the value of the cube in appropriate significant figures is (take 1 MSD = 1mm)

A
$$2.76 cm^{3}$$

B
$$1.03 cm^{3}$$

C
$$1.12 cm^{3}$$

D
$$1.00 cm^{3}$$

Solution

Least count of Vernier calipers
$$ 1 mm - \dfrac{8}{10} mm = 0.2 mm $$
Measured reading = $$ MSR + VSR (LC) = 10 mm + 2 (0.2 mm)$$
$$ = 10.4mm $$
Volume $$= (10.4)^3 mm^3$$
$$ V = 1.12 cm^3 $$
Question 8
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A uniform rod of length $$2l$$ is placed with one end in contact with the horizontal table and is then inclined at an angle $$\alpha $$ to the horizontal and allowed to fall. When it becomes horizontal, its angular velocity will be

A
$$ \omega = \sqrt{\left (\dfrac{3g \sin \alpha}{2l} \right )} $$

B
$$ \omega = \sqrt{\left (\dfrac{2l}{3g \sin \alpha} \right )} $$

C
$$ \omega = \sqrt{\left (\dfrac{g \sin \alpha}{l} \right )} $$

D
$$ \omega = \sqrt{\left (\dfrac{l}{g \sin \alpha} \right )} $$

Solution

$$mg \dfrac{l}{2} \sin \alpha = \dfrac{1}{2} I \omega^{2} $$

$$ \dfrac{mg \sin \alpha}{2} = \dfrac{1}{2} \dfrac{ml^2}{3} \omega^{2} $$

$$ \sqrt{\dfrac{3g}{l} \sin \alpha} = \omega $$

$$l$$ gives in $$2l$$
So $$ \omega = \sqrt{\dfrac{3g}{2l} \sin \alpha} $$
Question 9
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Find the acceleration of block of mass m. Assume pulleys are massless and frictionless.

A
$$g/3$$

B
$$2g/3$$

C
$$g/2$$

D
None of these

Solution

For $$2m,$$
$$2mg = 2T = 2m(\dfrac{a}{2})$$
for m,
$$T = ma$$
$$\therefore 2mg = 2(ma) = ma$$
$$a = \dfrac{2g}{3}$$
Question 10
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The angular speed of a body changes from $$\omega_{0}$$ to $$\omega,$$ solely due to a change in its moment of inertia. The ratio of the radii of gyration is

A
$$\omega : \omega_{0}$$

B
$$\sqrt{\omega} : \sqrt{\omega_0}$$

C
$$\sqrt[3]{\omega} : \sqrt[3]{\omega_0}$$

D
$$1 : 1$$

Solution

Since there is no external torque
So, angular momentum is conserved
$$L_{1} = L_{2}$$
$$I_{1}w_{1} = I_{2}w_{2}$$
$$w_{1} = w_{0}$$
$$w_{2} = w$$
$$\dfrac{I_1}{I_2} = \dfrac{w_2}{w_1} = \dfrac{w}{w_0} = w : w_0$$

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Re-Attempt Weekly Quiz Competition

Basic Science Test 26

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Basic Science Test 26

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Question 1

$$72$$ miles per hour

$$75$$ miles per hour

$$65$$ miles per hour

$$78$$ miles per hour

Solution

Question 2

$$g = 20\ ms^{-2} $$

$$g = 25\ ms^{-2} $$

$$g = 15\ ms^{-2}$$

$$g = 30\ ms^{-2} $$

Solution

Question 3

$$5\ cm $$

$$10\ cm $$

$$15\ cm $$

$$20\ cm $$

Solution

Question 4

$$1/ \omega_{2} : 1/\omega_{1} $$

$$\omega^{2}_{2} : \omega^{2}_{1} $$

$$ \sqrt{\omega_{2}} : \sqrt{\omega_{1}} $$

$$ \omega_{2} : \omega_{1} $$

Solution

Question 5

$$ \dfrac{g(t_{1} + t_{2})^{2}}{4} $$

$$ \dfrac{g(t_{1} + t_{2})^{2}}{8} $$

$$ \dfrac{3}{8} g (t_{1} + t_{2})^{2} $$

$$ \dfrac{g (t_{1} + t_{2})^{2}}{2} $$

Solution

Question 6

$$10\%$$

$$14\%$$

$$17\%$$

$$19\%$$

Solution

Question 7

$$2.76 cm^{3}$$

$$1.03 cm^{3}$$

$$1.12 cm^{3}$$

$$1.00 cm^{3}$$

Solution

Question 8

$$ \omega = \sqrt{\left (\dfrac{3g \sin \alpha}{2l} \right )} $$

$$ \omega = \sqrt{\left (\dfrac{2l}{3g \sin \alpha} \right )} $$

$$ \omega = \sqrt{\left (\dfrac{g \sin \alpha}{l} \right )} $$

$$ \omega = \sqrt{\left (\dfrac{l}{g \sin \alpha} \right )} $$

Solution

Question 9

$$g/3$$

$$2g/3$$

$$g/2$$

None of these

Solution

Question 10

$$\omega : \omega_{0}$$

$$\sqrt{\omega} : \sqrt{\omega_0}$$

$$\sqrt[3]{\omega} : \sqrt[3]{\omega_0}$$

$$1 : 1$$

Solution

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