MCQ Questions for Class 8 General Knowledge Basic Science Quiz 7

Which of the following instrument is the least precise device for measring length?

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Vernier callipers with 50 division of Vernier scale equal to 49 division of main scale and each division of main scale is of 1 mn.
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A screw scale having 100 divisions and each division is of 1 cm.
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A metre scale having 100 divisions and each division is of 1 cm.
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An optical instrument that can resolve upto $10^{-7} m$ .

Explanation

Most precise device is the one with the minimum least count

1. Vernier Calliper
20 divisions on the sliding scale

L.C=

$\frac{1}{20}mm$ = 0.05mm

2. Screw Gauge
pitch 1 mm and 100 divisions on the circular scale

L.C=

$\frac{1}{100}mm$ = 0.01mm

3. meter Scale
pitch 1 mm and 100 divisions on the scale

L.C=

$\frac{1}{100}mm$ =

$10^{-2}mm$

4.optical instrument
within a wavelength of light
Wavelenght of light $\approx$ $589 n m$

Hence option D is correct

A vernier callipers (with least count = 0 .1 mm ) has 20 divisions of the vernier scale .The main scale division are of

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0 .2 mm
0%
0.5 mm
0%
1.0 mm
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2.0 mm

The diameter of a disc is

$1m$ . It has a mass of

$20kg$ . It is rotating about its axis with a speed of

$120$ rotations in one minute. Its angular momentum in Kg

$m^2/s$ is

0%
$3.4$
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$31.4$
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$41.4$
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$43.4$

Explanation

We know that

$L=I\omega$

$L=\frac{Mr^2}{2}*2\pi n$

$=\frac{20*1^2}{2}+\frac{2*3.14*120}{60}$

$=31.4$

M gram of ice at

$0^{\circ} C$ is mixed with M gram of water at

$10^{\circ} C.$ The final temperature is

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$8^{\circ} C$
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$6^{\circ} C$
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$4^{\circ} C$
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$0^{\circ} C$

Explanation

Let final temperature = T then
Heat required to melt ice and to increase temperature by T

$H_1 = ML + MC\Delta T$

$T = \dfrac{-530}{2} = -265^{\circ} C$
But it can reach only

$0^{\circ} C.$
So partly the ice will melt and all the water will be at

$0^{\circ} C.$

A body takes

$5$ min to cool from

$80^\circ C$ to

$70^\circ C.$ To cool from

$80^\circ C$ to

$60^\circ C,$ it will take (Room temperature =

$40^\circ C$ )

0%
$5$ min
0%
$10$ min
0%
$12$ min
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$14$ min

Explanation

Newton's law of cooling, we have

$mc(T_1 - T_2) = k \left [ \dfrac{(T_1 + T_2)}{2} - T_0 \right ]$ .......(i)

$\dfrac{mc}{K} = \left [ \dfrac{(T_1 + T_2)}{2} - T_0 \right ] / \left [ \dfrac{(T_1 - T_2)}{t} \right ]$

$\dfrac{mc}{K} = \left [ \dfrac{80 + 70)}{2} - 40 \right ] / \left [ \dfrac{(80 - 70)}{5} \right ]$

Substituting

$T_{1} = 80$ &

$T_{2} = 60$ &

$T_{0} = 40$

equation (i)

$mc \dfrac{(80 - 60)}{t} = K \left [ \dfrac{(80 + 60)}{2} - 40 \right ]$

$20\dfrac{mc}{t} = k \times 30$

$t = \dfrac{20\,mc}{30\,K}$

$= \dfrac{20}{30} \times \dfrac{35}{2}$

$= \dfrac{70}{6}$

$t = 11.66 = 12\,min$

Choose the correct statements from the following. The range of a projectile depends upon

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The angle of projection
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The acceleration due to gravity
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The magnitude of the velocity of projection
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The mass of the projectile

Which of the statement are correct ?
(1) An electron

$4p$ orbital has more energy than one in

$3p$ orbital
(2) An electron in

$4s$ orbital is further from the nucleus than one in a

$3s$ orbital
(3) the angular quantum number

$1$ is related to orbital shape.
(4) Energy is released if electron moves from

$n=2$ to

$n=5$ level.

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$1,2$ and $3$
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$1$ and $3$
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$2$ and $4$
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Only $4$

Explanation

$n+l$ rule
1)

$4p=4+1$

$3p=3+1$

$4p>3p$
2) energy of

$4s>3s$ so

$4s$ is further
3) correct
4)

$2$ to

$5$ energy is absorbed

$x$

A wedge

$ABC$ and a block are placed as shown in the figure. The is no friction at any surface. Displacement of wedge

$ABC$ when

$2\ kg$ block reaches the point

$B$ :

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$1.6\ m$
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$2\ m$
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$1.5\ m$
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$1.2\ m$

Explanation

$\because$ No ant. force is present in horizontal direction,

$\Delta x_{cm}=0$

$\Rightarrow _1 \Delta x_1=m_2 \Delta x_2$

$\Rightarrow 2 \times (y-x)=8x$

$\Rightarrow y-x=4x$

$\Rightarrow 5x=y$

Now, from trigo

$\tan 37^o=\dfrac{6}{y}$

$\Rightarrow y=6\ cot 37^o$

$\Rightarrow x=\dfrac{6}{5} cot 37^o$

A hypothetical vernier scale of a travelling microscope has 40 divisions which is equal to 38 main scale divisions if each main scale division is 1.2 mm then minimum error in the measurement of length is

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$0.6 mm$
0%
$1.2 mm$
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$0.06 mm$
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$0.1 mm$

Explanation

We know that

$40 VSR = 3 SMS R$

$WSR = \frac { 38}{40}$

least count

$= 1 - \frac {38}{40} MSR$

$= \frac {2}{40} = \frac {1}{20} \times 1.2 = 0.06$

(i) Joule is the unit of force
(ii) Positively charged ion are called anions

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True
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False

Explanation

(I)

$\rightarrow$ Force

$\rightarrow$ Newton

Joule is unit of work

(Ii)

$\rightarrow$ anions

$\rightarrow$ negatively change

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

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An electron is emitted as a beta particle from an inner electron shell of the atom
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An electron is emitted as a beta particle from an outer electron shell of the atom
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A neutron changes into a proton and a beta particle is emitted from the nucleus
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A proton changes into a neutron and a beta particle is emitted from the nucleus

Explanation

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

$n\rightarrow p+\beta ^-+\vec v_{e}$

A body of mass

$5\ kg$ moving horizontally at a speed of

$1.5\ ms^{-1}$ . A perpendicular force of

$5\ N$ acts on it for

$4$ second distance of the body from the point of application of the force is

0%
$25\ m$
0%
$6\ m$
0%
$8\ m$
0%
$10\ m$

Explanation

$F=\dfrac {dm}{dt}v$

$=\dfrac {50}{1000}\times 500$

$=25 N$

A car moving with speed 30 m/s on circle of radius 500 m. Its speed is increasing at of

$2 m/s^{2}$ . The acceleration of the car is

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$9 .8 m/s ^{2}$
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$1 .8 m/s ^{2}$
0%
$2 m/s ^{2}$
0%
$2 .7 m/s ^{2}$

Explanation

Answer
Tangentical acceleration

$a_t= 2m/sec^{2}$

Velocity (v) = 30 m /sec
Radius (R) = 500 m

$(a_{c}) = \dfrac{v^{2}}{R} = \dfrac{(30)^{2}}{R}$

$= \dfrac{900}{500} = 1.8 m /sec{2}$

$= \sqrt{a_t^{2} + a_c^{2}}$

$= \sqrt{4 + 3. 24}$

$2 .7 m/sec^{2}$

$a_{net} = \sqrt{4 + 3.24}$

$a_{net}= \sqrt{7.24}$

$a_{net}= 2. 99 = 2.7$

The moment of inertia of a solid sphere , about an axis parallel to its diameter and at a distance of a from it is . , I ' (x) ' which one of the graph represents the variation of I ( x ) with x correctly

The velocity - time graph for a particle travelling along a straight line as shown in figure , if the graph is a sine curve , the average velocity during the first 5 s is 5.8 m /ms

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6.37 m /s
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2.37 m/s
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2.37 m/s
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7.5 m/s

The gravitational force in region given by

$\vec{F} = a y \hat{i} + ax \hat{j}$ . The work done by gravitational force to shift a point ,mass m from ( 0 , 0 , 0) to

$(x_{0}, y_{0},z_{0})$ is

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$max_{0}y_{0}z_{0}$
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$max_{0}y_{0}$
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$- max_{0}y_{0}$
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0

The moment of inertia of a wheel is

$100 kg - m ^{2}$ . At a given instant , its angular velocity is 10 rad /s . after the wheel rotates through an angle of 100 radians the wheel 's angular velocity is 100 rad /s . The torque applied on wheel is

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$4.95 \times 10^{5} N. m$
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$4.95 \times 10^{4} N. m$
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$4.95 \times 10^{3} N. m$
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$49.5 \times 10^{5} N. m$

A body is dropped from a balloon moving up with a velocity of

$4\ ms^{-1}$ when the balloon is at a height of

$120.5\ m$ from the ground; the height of the body after five seconds from the ground is :

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$8\ m$
0%
$12\ m$
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$18\ m$
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$24\ m$

Explanation

Using second equation of newton

$S = ut + \dfrac{1}{2} at^{2}$

$= 4 \times 5 + \dfrac{1}{2} \times 9.8 \times (5)^{2}$

$= 20 + \dfrac{1}{2} \times 9.8 \times 25$

$= 20 + 82.5$

$=102.5$

Hence present height

$= 120.5 - 102.5$

$= 18\ m$

The maximum positive displacement

$x$ is

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$2\sqrt{3}\ m$
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$2\ m$
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$4\ m$
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$\sqrt{2}\ m$

A square metal frame is made of four identical thin rods each of the mass 6 kg and lengthm. The moment of inertial of the frame is

$kg - m ^{2}$ about an axis passing through point O and perpendicular to the plane of the frame is

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4
0%
8
0%
12
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1

The unit of expression

$\mu_{0} \varepsilon_{0}$ are :-

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$m/s$
0%
$m^{2} /s^{2}$
0%
$s^{2} / m^{2}$
0%
$s/m$

Explanation

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

$C$ - Speed of light

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

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

$= \dfrac{S^2}{m^2}$

Consider the situation as shown in the figure. The system is free to move.
The force exerted by the rope on the block is FM/(m + m)

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only if the rope is uniform
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in gravity free space
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only if F > (M + m)g
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in all cases

Figure given below shows the graph of velocity

$v$ of particle moving along

$x-axis$ as a function of time

$t$ . Average acceleration during

$t = 1\ s$ to

$t = 7\ s$ is

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$1.5 m/s^{2}$
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$1 m/s^{2}$
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$2 m/s^{2}$
0%
$2.5 m/s^{2}$

Explanation

$a = \frac {dv}{dt} =$ solpe of v-f

$\frac { 20 -0}{2 - 0} = 10 m/s^2 0-2$

$2- 6$

$\frac {30-20}{8-6} = \frac {10}{2} = 5 m/s^2 = 6.8$
Average acc

$\Rightarrow \frac { 10 \times 1+ 0 \times 4 +5 \times 1}{6}$

$=\frac {15}{6} = 2.5$

1874420_1901198_ans_c3400db488ab4979b1d6fd33e5f8c954.png

a stationary where starts roatating about its own axis at constant angular acceleration if the wheel completes 50 rotatations made by it in next two seconds is

0%
$75$
0%
$100$
0%
$125$
0%
$150$

Explanation

$\theta = cot + \frac {1}{2} \alpha t^2$

$100 M = \frac {1}{2} \times (2)^2$

$\alpha = 50 \pi ral /s^2$

$\therefore \theta$ for 2 next / second

$= 40 com - 1 com$

$= 300 \pi$

$\therefore$ no f revolute =

$\frac { 300 M}{2M} = 150$

A metal tape gives correct measurement at

$15^{\circ}C$ . It is used to measure a distance of

$100m$ at

$45^{\circ}C$ . The error in the measurement, if

$\alpha = 12\times 10^{-6}/1^{\circ}C$ is

0%
$36\ cm$
0%
$36\ m$
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$42\ mm$
0%
$36\ mm$

Explanation

Correct measurement at

$15^{\circ}C$

$\% error = \dfrac {\triangle l}{l}\times 100$

$= \dfrac {l\alpha \triangle T}{l} \times 100$

$\dfrac {\triangle l}{l} \times 100 = 12\times 10^{-6} \times (45 - 15) \times 100$

$\triangle l = l\times 12\times 10^{-6} \times 30$

$= 100\times 12\times 10^{-6} \times 30$

$= 360\times 10^{-6} = 36\ mm$ .

How many ATP molecules are formed from one molecule of acetyl

$CoA$ in

$TCA$ cycle

0%
$1$
0%
$13$
0%
$24$
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$2$

Two equally charged identical pith balls are suspended by identical massless strings as shown in the adjacent figure if this set up is mercury

$( g = 3.7 m/s^2)earth ( g = 9.8 m/s^2)$ and

$Jupiter ( g =m/s^2)$ then angle

$2 \theta$ will be

0%
maximum on mercury
0%
Maximum on earth as it ha atmosphere
0%
maximum on jupiter
0%
the same on any planet as coulomb force is independent of gravity

Explanation

$tan \theta = \frac {F}{ mg} = \frac { \frac { kq^2}{ 4 sin^2 \theta } }{ mg}$

$= \frac { kq^2}{ 4 sin^2 \theta mg }$

$tan \theta \times in^2 \theta = \frac {kq^2}{ 4 m g }$

$\frac { sin^3 \theta }{ cos \theta } = \frac { ka^2}{4 m} ( \frac {1}{g} )$

$tan^2 \theta = sin \theta \alpha \frac {1}{g}$

$\theta \alpha \frac {1}{g}$
So

$\theta$ is max when g is min so g is on mercury
1874411_1901184_ans_72147b105117470681084b6222bdfd8b.png

Two particle A and B are thrown vertically upward simultaneously , with velocity 5 m/s and 15 m/s respectively from the same position the separation between them after 1 s is

$( g = 10 m/s^2)$

0%
$7.5 m$
0%
$5 m$
0%
$10 m$
0%
$Zero$

Explanation

$S_1 - \mu t - l_2 gl^2$

$= 5 \times 1 - \frac {1}{2} \times 10 l^2$

$= S - S$

$= OM$

$S_2 = \mu_2 t - t_2 gl^2$

$= 15 \times 1 - \frac {1}{2} \times 10$

$= 15 - 5 = 10 m$

$\Delta S = S_2 - S_1 = 10 - 0 = 10 m$
1874422_1901201_ans_45a192ce73a245be9340a4f0b7b7913c.png

A particle ( 5 kg) moves upon straight line with constant velocity

$2 ms^{-1}$ along the straight line

$2y = 3 x+ 4$ the angular momentum of particle about origin in

$kg m^2s^{-1}$ will be

0%
$\frac {30}{ \sqrt { 13} }$
0%
$\frac { 40 }{ \sqrt {13} }$
0%
$\frac {20}{ \sqrt { 13} }$
0%
$\frac {10}{ \sqrt {13} }$

Explanation

Angular momentum

$|Rightarrow m ( \overrightarrow {r} \times \overrightarrow {V} ) = m \mu r$
so velocity using x
direction will be

$= 2 \times cos \theta$

$\Rightarrow 2 \times \frac {2}{ \sqrt {3}}= \frac {4}{ \sqrt { 13} } m/s$
Velocity along y direction

$\Rightarrow 2 sin \theta = 2 \times \frac {3}{ \sqrt {13} }$

$\Rightarrow \frac {6}{ \sqrt {13} } \hat {j}$ m/s
magnitude of nel = 2m/s
distance of origin to the line

$y = \frac {3}{2} \times 12$

$|d| = | \frac { 0 +0 -2}{ \sqrt { 1+ 9/4} } | \Rightarrow \frac { 2 \times 2 }{ \sqrt {13} } = \frac {y }{ \sqrt {13} }$
Angular momentum =

$m \mu r$

$5 \times 2 \times \frac {4}{ \sqrt {13} } = \frac {40}{ \sqrt {13}} kg \frac {m^2}{s}$
1874395_1901180_ans_cb9774aafc954147b83a465d6da50fc2.png

Two small blocks A of mass

$\sqrt {3} kg$ and

$B$ of mass kg are connected by a light string are placed on a smooth fixed sphere such that the system is in equilibrium for

$< AOB = \frac { \pi}{2}$ then

0%
The value of $\alpha = \frac { \pi}{3}$ and equilibrium is table
0%
The value of $\alpha = \frac { \pi}{3}$ and equilibrium is unstable
0%
The value of $\alpha = \frac { \pi}{6}$ and equilibrium is stable
0%
The value of $\alpha = \frac {\pi}{6}$ and equilibrium is unstable

Explanation

$T = mg sin \alpha$

$T = mg sin ( so \div \alpha)$

$\Rightarrow ma = mg$

$\therefore g sin \alpha = \sqrt {3} g cos \alpha$

$fan \alpha = \sqrt {3} 2 tan \frac { \pi}{3}$

$\therefore \alpha = \frac { \pi}{3}$ at stable
1874060_1901263_ans_4bd17afb103b4854b295de67f1c1a27c.png

CBSE Questions for Class 8 General Knowledge Basic Science Quiz 7 - MCQExams.com

Most precise device is the one with the minimum least count

1. Vernier Calliper
20 divisions on the sliding scale

2. Screw Gauge
pitch 1 mm and 100 divisions on the circular scale

3. meter Scale
pitch 1 mm and 100 divisions on the scale

4.optical instrument
within a wavelength of light
Wavelenght of light $\approx$ $589 n m$

0:0:1

Practice Class 8 General Knowledge Quiz Questions and Answers

CBSE Questions for Class 8 General Knowledge Basic Science Quiz 7 - MCQExams.com

Most precise device is the one with the minimum least count

1. Vernier Calliper20 divisions on the sliding scale

2. Screw Gaugepitch 1 mm and 100 divisions on the circular scale

3. meter Scalepitch 1 mm and 100 divisions on the scale

4.optical instrumentwithin a wavelength of lightWavelenght of light ≈\approx 589nm

0:0:1

Practice Class 8 General Knowledge Quiz Questions and Answers

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1. Vernier Calliper
20 divisions on the sliding scale

2. Screw Gauge
pitch 1 mm and 100 divisions on the circular scale

3. meter Scale
pitch 1 mm and 100 divisions on the scale

4.optical instrument
within a wavelength of light
Wavelenght of light $\approx$ $589 n m$