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CBSE Questions for Class 8 General Knowledge Basic Science Quiz 10 - MCQExams.com
CBSE
Class 8 General Knowledge
Basic Science
Quiz 10
Which of the following could not be a unit of acceleration?
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$$km/s^2$$
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$$cm s^{-2}$$
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$$ km/s$$
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$$ m/s^2$$
Explanation
$$ km/s$$
The SI unit of measuring momentum of a moving body is
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$$ms^1$$
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$$kg/ms$$
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$$kgms^{-1}$$
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$$kgm's$$
Explanation
(C) $$kgms^{-1}$$
A component pipeline must perform unit tests.
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True
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False
Explanation
True. A component pipeline must perform unit tests.
Unit testing is a software development and testing process in which the testable parts of an application known as units are tested individually and independently to check if they are working as per the requirement.
This testing can be done manually but is generally programmed.
It is a part of Test Driven Development methodology.
In unit testing the developer first write failing unit tests.
Then the code is written to make the changes in the application until it passes the test.
$$1$$ amu is
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$$1.66 \times 10^{-24}kg$$
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$$1.66 \times 10^{-27}kg$$
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$$\dfrac{1}{NA}$$
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Both (B) and (C)
Explanation
$$1$$ amu = $$1.66 \times 10^{-24}gm=1.66\times 10^{-27}kg$$
Weight of $$1$$ element or atom
$$1$$ mole = $$N_A$$ atom = Atomic mass
So, weight of one atom = $$(\dfrac{1}{N_A})$$ called amu.
Two light strings of length $$4cm$$ and $$3cm$$ are tied to a bob of weight $$500gm$$. The free ends of the strings are tied to pegs in the same horizontal line and separated by $$5cm$$. The ratio of tension in the longer string to that in the shorter string is
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$$4:3$$
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$$3:4$$
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$$4:5$$
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$$5:4$$
Explanation
Let the two pegs be 'B' and 'C' and the bob be at 'A'.
Let the tension in the 2 strings be $$T_1$$ and $$T_2$$
$$T_1\cos\theta=T_2\sin\theta$$
$$\frac{T_1}{T_2}=\cot\theta=\frac{3}{4}$$
A bob is suspended from a ideal string of length 'l'. Now it is pulled to a side through $$60^0$$ to vertical and whirled along a horizontal circle. Then its period of revolution is
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$$\pi\sqrt{l/g}$$
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$$\pi\sqrt{l/2g}$$
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$$\pi\sqrt{2l/g}$$
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$$2\pi\sqrt{l/g}$$
Explanation
From F.D
$$T=60^0=mg$$
$$T*\frac{1}{2}=mg$$
$$T=2mg$$....(i)
$$\sin60^0=mdw^2$$
$$mg*\frac{\sqrt{3}}{2}=m*l\sin60^0*\frac{4\pi^2}{T^2}$$
$$T^2=l\frac{4\pi^2}{2g}$$
$$T=\pi\sqrt{\frac{2l}{g}}$$
The direction of projection of particle is shown in the figure for an observer on trolley. An observer on the ground sees the ball rise vertically. The maximum height reached by ball as seen from the trolley is
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$$10m$$
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$$15m,$$
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$$20m$$
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$$5m$$
Explanation
Horizontal V = 0
Vcos60 - 10=0
V x 1/2 =10
V = 20m/s
Vertical Velocity = Vsin60
20 x ((3)^-1/2)/2) = 10((3)^1/2) m/s
2ay = v^2 -u^2
2 x (-10) x h = 0 - 1
0((3)^1/2)^2
-20h = -300
h= 15m
OR
$$10-v\cos60^0=0$$
$$v=20m/s$$
$$H=\frac{v^2\sin^260^0}{2g}=15m$$
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
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$$10\%$$
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$$14\%$$
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$$17\%$$
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$$19\%$$
Explanation
$$\begin{array}{r} \% \dfrac{\Delta x}{x}=\left(3 \% \dfrac{\Delta a}{a}\right)+\left(2 \% \dfrac{\Delta b}{b}\right)+\left(\% \dfrac{\Delta d}{d}\right) +\left(\dfrac{1}{2} \% \dfrac{\Delta c}{c}\right) \end{array}$$
Given that
$$\%\dfrac{\Delta a}{a}=1\,\,\,\,\,\,\%\dfrac{\Delta b}{b}=2$$
$$\%\dfrac{\Delta c}{c}=3\,\,\,\,\,\,\, \%\dfrac{\Delta d}{d}=4$$
$$\dfrac{\Delta x}{x}=3 \times 1+2 \times 3+4+\dfrac{1}{2} \times 2$$
$$\dfrac{\Delta x}{x}=14\%$$
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)
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$$2.76 cm^{3}$$
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$$1.03 cm^{3}$$
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$$1.12 cm^{3}$$
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$$1.00 cm^{3}$$
Explanation
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 $$= (10.4)^3 mm^3$$
$$ V = 1.12 cm^3 $$
Determine force on P in the following system of the charges q kept vertices of an equilateral triangle of side 'a'.
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$$ \sqrt{2} \dfrac{kq^2}{a^2} $$
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$$ \sqrt{3} \dfrac{kq^2}{a^2} $$
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$$ \dfrac{kq^2}{a^2} $$
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$$ \dfrac{4kq^2}{a^2} $$
Explanation
$$F_{net} = \sqrt{F^{2} + F^{2} + 2 F^{2} \cos 60} $$
$$ = \sqrt{F^{2} + F^{2} + F^{2}} $$
$$ = \sqrt{3} F $$
$$ = \sqrt{3} \dfrac{kq^{2}}{a^2} $$
The dimensions of Planck's constant is
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$${M^2L^2T^{-1}}$$
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$${M^2LT^{-2}}$$
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$${ML^2T^{-1}}$$
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$${ML^2T^{-2}}$$
A particle starts from rest at t = 0 and moves along a straight line with a time varying acceleration giv en by $$ a (t) = a _{0} a ^{-bt} $$ , where $$a_{0} $$ and b are positive contains . Its velocity (v) versus time (t) graph is best given by
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Choose the correct answer from the alternatives given.
Time period is the time taken to complete one oscillation. This definition of time period is true for
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longitudinal Wave
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transverse Wave
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simple Pendulum
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All the above
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