Units and Measurements Test

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Units and Measurements Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

When $$24.25\times { 10 }^{ 3 }$$ is rounded off to three significant figures, it becomes :

A
242

B
243

C
$$244\times { 10 }^{ 2 }$$

D
$$24.2\times { 10 }^{ 3 }$$

Solution

$$24.25\times 10^{3} \rightarrow 2,4,2,5\rightarrow $$ present by it has 4 significant digits.
If the digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if it is even.
So, up to three significant figures, the number $$=24.2\times 10^{3}$$
Question 2
1 / -0

Universal gravitational constant is given by $$6.67\times { 10 }^{ -11 }N{ m }^{ 2 }{ kg }^{ -2 }$$. Then the no. of significant figures in it is:

A
$$14$$

B
$$2$$

C
$$3$$

D
$$11$$

Solution

$$6.67\times 10^{-11}$$ has $$3$$ significant digits $$\rightarrow6,6,7$$
Question 3
1 / -0

The radius of a sphere is $$5$$ cm. Its volume will be given by (according to the theory of significant figures) :

A
$$523.33\ { cm }^{ 3 }$$

B
$$5.23\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$$

C
$$5.0\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$$

D
$$5\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$$

Solution

Volume$$=\dfrac{4}{3}{\pi r}^{3}=\dfrac{4}{3}\times \pi \left ( 5 \right )^{3}=523.33\ {cm}^{3}$$$$=5.2333\times 10^{2}\ cm^{3}$$
$$\downarrow $$
5 significant figures
Since radius has single significant figure, so, volume should also have single significant figure.
For single significant figure, we have to drop all after decimal.
$$\Rightarrow$$ Volume $$= 5\times 10^{2}\ cm^{2}$$ (if the digit to be dropped is less than 5, preceding digit is left unchanged)
Question 4
1 / -0

Two numbers are given as i) 20.96 and ii) 0.0003125. Rounding off above numbers to 3 significant figures will lead to :

A
i) 21.0, ii) 312

B
i) 21.0, ii) 3.12 $$\times{ 10 }^{ -4 }$$

C
i) 2.10, ii) 3.12 $$\times{ 10 }^{ -4 }$$

D
i) 210, ii) 3.12 $$\times{ 10 }^{ -4 }$$

Solution

a) 20.96
We have to round off for 3 significant digits. Last digit has to be rounded off.
If the digit to be dropped (6) is more than 5, then preceding digit is raised by 1.
Hence $$20.96\approx 21.0$$

b) $$0.0003125=3.125\times 10^{-4}$$
For 3 significant digits $$\Rightarrow 3.12\times 10^{-4}$$
Question 5
1 / -0

The dimension of time in Electric field intensity is :

A
$$-1$$

B
$$-2$$

C
$$-3$$

D
$$3$$

Solution

$$F=QE$$
$$E=F/Q$$

$$\Rightarrow \dfrac{MLT^{-2}}{AT}=MLT^{-3}A^{-1}$$
Question 6
1 / -0

When we express the velocity of light, $$30,00,00,000/ ms^{-1}$$, in standard form up to three significant figures, it's value is:

A
$$3\times { 10 }^{ 8 }\ { ms }^{ -1 }$$

B
$$3.00\times { 10 }^{ 8 }\ { ms }^{ -1 }$$

C
$$3\times { 10 }^{ 10}\ { cms }^{ -1 }$$

D
$$3\times { 10 }^{ 6}\ { ms }^{ -1 }$$

Solution

Since all the zeros on the right of last non zero digit in the decimal part are significant.
$$\Rightarrow 300000000=3.00\times 10^{8}\ m/s$$
Question 7
1 / -0

If $$'m$$' is the mass of a body, $$'a'$$ is amplitude of vibration, and $$'w '$$ is the angular frequency then $$\frac { 1 }{ 2 } { ma }^{ 2 }{ w }^{ 2 }$$ has same dimensional formula as :

A
work

B
moment of force

C
energy

D
all the above

Solution

Dimension of $$\dfrac{1}{2}ma^{2}w^{2}=ML^{2}\left ( \dfrac{1}{T^{2}} \right )=ML^{2}T^{-2}$$
Work$$=F.d\left ( force\times distance \right )=\left ( \dfrac{ML}{T^{2}} \right )\times L=ML^{2}T^{-2}$$
Moment of force $$=\underset{f}{\rightarrow}\times \underset{r}{\rightarrow}=\left [ \dfrac{ML}{T^{2}} \right ]\times L=ML^{2}T^{-2}$$
Energy = work = $$ML^{2}T^{-2}$$
Question 8
1 / -0

$$\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -3 }{ A }^{ -2 } \right] $$is the dimensional formula of :

A
Electric resistance

B
Capacity

C
Electric potential

D
Specific resistance

Solution

$$Power=I^2R$$
$$[R]=\dfrac{ML^2T^{-3}}{A^2}$$
Option A is correct.
Question 9
1 / -0

The dimension of magnetic induction in $$M, L, T$$ and $$C$$ (Coloumb) is given as:

A
$${ MT }^{ -1 }{ C }^{ -1 }$$

B
$${ MT }^{ -2 }{ C }^{ -1 }$$

C
$${ MLT }^{ -1 }{ C }^{ -1 }$$

D
$${ MT }^{ 2 }{ C }^{ -2 }$$

Solution

$$Force = Bil = M^{1}L^{1}T^{-2}$$

$$ B = \dfrac{M^{1}L^{1}T^{-2}}{I^{1}L^{1}} = M^{1}I^{-1}T^{-2}$$

$$C = Charge = I^{1}T^{1}$$

$$So, B = M^{1}C^{-1}T^{-1}$$
Question 10
1 / -0

The ratio $$\dfrac { L }{ R } $$ has the dimensions of:
[ L:Inductance R:Resistance]

A
Velocity

B
Acceleration

C
Time

D
Force

Solution

$$L$$ has the diemnsions of $$ML^2T^{-2}I^{-2}$$.
Resistance has the dimensions $$ML^2T^{-3}I^{-2}$$.
$$\dfrac{L}{R}=\dfrac{ML^2T^{-2}I^{-2}}{ML^2T^{-3}I^{-2}}=T$$

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Re-Attempt Weekly Quiz Competition

Units and Measurements Test - 24

Result

Units and Measurements Test - 24

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SHARING IS CARING

Question 1

242

243

$$244\times { 10 }^{ 2 }$$

$$24.2\times { 10 }^{ 3 }$$

Solution

Question 2

$$14$$

$$2$$

$$3$$

$$11$$

Solution

Question 3

$$523.33\ { cm }^{ 3 }$$

$$5.23\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$$

$$5.0\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$$

$$5\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$$

Solution

Question 4

i) 21.0, ii) 312

i) 21.0, ii) 3.12 $$\times{ 10 }^{ -4 }$$

i) 2.10, ii) 3.12 $$\times{ 10 }^{ -4 }$$

i) 210, ii) 3.12 $$\times{ 10 }^{ -4 }$$

Solution

Question 5

$$-1$$

$$-2$$

$$-3$$

$$3$$

Solution

Question 6

$$3\times { 10 }^{ 8 }\ { ms }^{ -1 }$$

$$3.00\times { 10 }^{ 8 }\ { ms }^{ -1 }$$

$$3\times { 10 }^{ 10}\ { cms }^{ -1 }$$

$$3\times { 10 }^{ 6}\ { ms }^{ -1 }$$

Solution

Question 7

work

moment of force

energy

all the above

Solution

Question 8

Electric resistance

Capacity

Electric potential

Specific resistance

Solution

Question 9

$${ MT }^{ -1 }{ C }^{ -1 }$$

$${ MT }^{ -2 }{ C }^{ -1 }$$

$${ MLT }^{ -1 }{ C }^{ -1 }$$

$${ MT }^{ 2 }{ C }^{ -2 }$$

Solution

Question 10

Velocity

Acceleration

Time

Force

Solution

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