Basic Science Test 18

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Basic Science Test 18

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Weekly Quiz Competition

Question 1
1 / -0

A beta particle is a fast-moving electron. Which statement explains how beta particles are emitted from an atom?

A
An electron is emitted as a beta particle from an inner electron shell of the atom

B
An electron is emitted as a beta particle from an outer electron shell of the atom

C
A neutron changes into a proton and a beta particle is emitted from the nucleus

D
A proton changes into a neutron and a beta particle is emitted from the nucleus

Solution

$$\therefore $$ The emission of beta particles is from the nucleus due to the nucleus conversion:

$$n\rightarrow p+\beta ^-+\vec v_{e}$$
Question 2
1 / -0

The unit of expression $$\mu_{0} \varepsilon_{0} $$ are :-

A
$$m/s$$

B
$$m^{2} /s^{2} $$

C
$$s^{2} / m^{2} $$

D
$$s/m$$

Solution

$$C = \dfrac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} $$
$$C$$ - Speed of light

$$ \sqrt{\mu_{0} \varepsilon_{0}} = \dfrac{1}{C}$$

$$ \mu_{0} \varepsilon_{0} = \dfrac{1}{C^2} = \dfrac{1}{\left (\dfrac{m}{s} \right )^{2}} $$

$$ = \dfrac{S^2}{m^2} $$
Question 3
1 / -0

A disc is rotating about its own axis with angular frequency 100 rpm. When a mass of 100 g is placed at 10 cm from its centre then its angular frequency becomes 50 rpm. The moment of inertia of the disc is-

A
$$5\times10^4 g cm^2$$

B
$$10^4 g cm^2$$

C
$$10^3 g cm^2$$

D
$$10^5 g cm^2$$
Question 4
1 / -0

The least count of stop watch is $$\dfrac{1}{5}$$ second.The time of $$20$$ oscillations of a pendulum is measured to be $$25$$ seconds. The maximum percentage error in the measurement of time will be

A
$$0.1\%$$

B
$$0.8\%$$

C
$$1.8\%$$

D
$$8\%$$
Question 5
1 / -0

A rod of length L and mass M is bent to semicircular ring as shown in the following figure the moment of inertia about XY is

A
$$ \frac {ML^2}{2 \pi^2 } $$

B
$$ \frac {ML^2}{ \pi^2} $$

C
$$ \frac {ML^2}{ 4 \pi^2 } $$

D
$$ \frac { 2ML^2}{ \pi^2 } $$

Solution

given,
Rod length L turns into semicricle of radius r
$$ L = \pi r $$
$$ r = \frac {L}{ \pi } $$
Inertia of ring $$ I = Mr^2 = \frac {ML^2}{ \pi^2 } $$
Hence , moment of inertia is $$ \frac {ML^2}{ \pi^2 } $$
Question 6
1 / -0

A researcher found petrified dinosaur faces. Which one of the following is unlikely to be found in this fossil?

A
Decayed conifer wood

B
Bamboo

C
Cyead

D
Gian tfern
Question 7
1 / -0

A crate is initially at rest on a horizontal frictionless table A constant horizontal force F is applied which of the following four graphs is a correct plot of work W as a function of the crate's speed V ?

A

B

C

D

Solution

Work energy relationship i.e. $$ W = \dfrac {1}{2} mu^2 $$
Indicates that the graph between W and u should be a parabola bent towards W axis
Question 8
1 / -0

The acceleration of a particle oscillating simple harmonically at distance $$ x_1 $$ and $$ x_2 $$ from the mean position are $$ a_1$$ and $$ a_2 $$ respectively then the period of oscillation is

A
$$ 2 \pi \sqrt { \frac {x_2 - x_1}{a_2 -a_1} } $$

B
$$ 2 \pi \sqrt { \frac {x_1 - x_2}{a_2 -a_1} } $$

C
$$ 2 \pi \sqrt { \frac {a_1 - a_2}{x_1 -x_2} } $$

D
none of these

Solution

We know that for SHM bmotion
$$ \frac {x^2_1}{ a^2} + \frac {a^2_1}{a^2 \omega^2} = \frac {x^2_2}{a^2} + \frac {a^2_2}{a^2 \omega^2 } $$
$$ x^2_1 - x^2_2 = ( a^2_ 2 - a^2_1 ) \frac {1}{ \omega^2} $$
$$ \omega = \sqrt { \frac {a^2_2 -a^2_1}{x^2_1 -x^2_2}} $$
and we know that
$$ \omega = \frac { 2 \pi}{ T } $$
$$T= $$ \sqrt { \frac{x^2_1 -x^2_2}{a^2_2 -a^2_1}}$$
Hence the correct option is D.
Question 9
1 / -0

$$I$$ is moment of inertia of a thin circular ring about an axis perpendicular to the plane of ring and passing through its centre. The same ring is folded into 2 turns coil. The moment of inertia of circular coil about an axis perpendicular to the plane of coil and passing through its centre is

A
2I

B
4I

C
$$\frac{I} {2}$$

D
$$\frac{I} {4}$$
Question 10
1 / -0

In a new system of units, the unit of mass equals $$\alpha$$ kg, the unit of length equals $$\beta$$ m and unit of time equals $$\gamma$$ second. The value of 1 newton in this new system of units is

A
$$\dfrac{\alpha\beta}{\gamma^2}$$

B
$$\dfrac{\alpha}{\beta\gamma^2}$$

C
$$\dfrac{\gamma}{\alpha\beta^2}$$

D
$$\dfrac{\gamma^2}{\alpha\beta}$$