M.C.Q (1 Marks)

MCQ 1011 Mark

The length of a rod is $(11.05\pm0.05)\text{cm}.$ What is the total length of $2$ such rods?

A
$(22.1\pm0.05)\text{cm}$
B
$(22.10\pm0.05)\text{cm}$
C
$(22.1\pm0.011)\text{cm}$
✓
$(22.1\pm0.10)\text{cm}$

Answer

Correct option: D.

$(22.1\pm0.10)\text{cm}$

MCQ 1021 Mark

If the length of a rectangle l = 10.5cm, breadth b= 2.1cm and minimum possible measurement by scale = 0.1cm, then the area is:

✓
22.0cm$^2$
B
21.1cm$^2$
C
22.05cm$^2$
D
22cm$^2$

Answer

Correct option: A.

22.0cm$^2$

Area of rectangle, A = Length × Breadth A = lb = 10.5 × 2.1= 22.05cm$^2$ Minimum possible measurement of scale = 0.1cm. So, area measured by scale = 22.0cm$^2$.

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MCQ 1031 Mark

In the formula $x = 3 yz^2, x$ and $z$ have dimensions of capacitance and magnetic induction, respectively. The dimensions of $y$ in $\ce{MKS}$ system are:

A
$[\text{M}^{-2}\text{L}^{-2}\text{T}^{4}\text{A}^{4}]$
B
$[\text{M}^{-3}\text{L}^{-3}\text{T}^4\text{A}^5]$
✓
$[\text{M}^{-3}\text{L}^{-2}\text{T}^8\text{A}^4]$
D
$[\text{M}^{-1}\text{L}^{-4}\text{T}^2\text{A}^4]$

Answer

Correct option: C.

$[\text{M}^{-3}\text{L}^{-2}\text{T}^8\text{A}^4]$

Given, $[x] =$ capacitance $=[\text{M}^{-1}\text{L}^{-2}\text{T}^4\text{A}^2]$
[Z] = magnetic induction $[\text{MA}^{-1}\text{T}^{-2}]$
So, $[\text{y}]=\frac{[\text{M}^{-1}\text{L}^{-2}\text{T}^{4}\text{A}^2]}{[\text{MA}^{-1}\text{T}^{-2}]^2}=[\text{M}^{-3}\text{L}^{-2}\text{T}^{8}\text{A}^{4}]$

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MCQ 1041 Mark

The ratio of the volume of the atom to the volume of the nucleus is of the order of:

✓
$10^{15}$
B
$10^{25}$
C
$10^{20}$
D
$10^{10}$

Answer

Correct option: A.

$10^{15}$

Radius of atom = $10^{-10}$m Radius of nucleus = $10^{-15}$m $\text{Ratio}=\frac{10^{-10}}{10^{-15}}=10^5$ Ratio of volume $=(10^5)^3=10^{15}$

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Physics M.C.Q (1 Marks)