Physics M.C.Q (1 Marks)

4. $[\text{ML}^2\text{T}^{-2}\text{k}^{-1}]$

1. Henry
2. Volt - Second/ Ampere

4. Joule/ ampere²

Explanation:

We know that $\frac{\text{L}}{\text{R}}=\text{t}\text{ and }\text{RC}=\text{t}$

$\frac{\text{L}}{\text{R}}=\text{RC}\text{ or }\text{CR}^2=\text{L}$

Unit of CR² are the same as unit of L, which is henry.

From $\text{e}=\text{L}=\frac{\text{edt}}{\text{dI}}=\frac{\text{Volt sec}}{\text{Ampere}}$

$\text{​​​​From}​​\text{U}=\frac{1}{2}\text{LI}^2,\text{L}=\frac{2\text{U}}{\text{I}^2}=\frac{\text{Joule}}{\text{Ampere}^2}$

4. An atomic clock.

Explanation:

The least count of a wall clock, stop watch, digital watch and atomic clock are 1 sec, $\frac{1}{10}\text{sec},\frac{1}{100}\text{sec and}\ \frac{1}{10^{13}}\text{sec}$ respectively. So atomic clock is most precise.

2. 10³ dyne

Explanation:

As, dimensional formula of force = [MLT^-2]

n₁ = 36, M₁ = 1kg, L₁ = 1m, T₁ = 1min = 60s

n₂ = ?, M₂ = 1g, L₂ = 1cm, T₂ = 1s

So, conversion of 36 units into CGS system

$\text{n}_2=\text{n}_1\Big[\frac{\text{M}_1}{\text{M}_2}\Big]^{\text{a}}\Big[\frac{\text{L}_1}{\text{L}_2}\Big]^{\text{b}}\Big[\frac{\text{T}_1}{\text{T}_2}\Big]^{\text{c}}$

$\text{n}_2=\text{n}_1\Big[\frac{\text{1Kg}}{\text{1g}}\Big]^{\text{1}}\Big[\frac{\text{1m}}{\text{1cm}}\Big]^{\text{1}}\Big[\frac{\text{1 min}}{\text{1 s}}\Big]^{\text{c}}$

$=36\Big[\frac{1000​​\text{g}}{1\text{g}}\Big]\Big[\frac{100\text{cm}}{1\text{cm}}\Big]^1\Big[\frac{60\text{s}}{1\text{s}}\Big]^{-2}=10^3\text{ dyne}$

1. $\frac{\text{R}}{\text{L}}$

1. $\frac{(\text{P}-\text{Q})}{\text{R}}$

5. $\frac{(\text{R}+\text{Q})}{\text{P}}$

Explanation:

In option (a) and (e) there is term (P - Q) and (R + Q) as different physical quantities can never be added or subtracted so option (a) and (e) can never be meaningful.

In option (b), the dimension of PQ may be equal to dimension of R so option (b) can be possible. Similarly dimensions of PR and Q² may be equal and gives the possibility of option (d).

In option (c), there is no addition subtraction gives the possibilities of option (c).

Hence, verifies the right option (a) and (e).

1. $\text{M}^{-1}\text{L}^3\text{T}^{-2}$

Explanation:

From $\text{F}=\frac{\text{G}\text{m}_1\text{m}_2}{\text{r}^2}$

$\text{G}=\frac{\text{F}\text{r}^2}{\text{m}_1\text{m}_2}=\frac{(\text{MLT}^{-2})\text{L}^2}{\text{M}^2}=[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$​​​​​​​

1. Resistance.

Explanation:

$\text{R}=\text{M}^1\text{L}^2\text{T}^{-3}\text{A}^{-2}\dots$ from

$​​\text{R}=\frac{\text{V}}{\text{I}}=\frac{\frac{\text{W}}{\text{q}}}{\text{I}}=\frac{\text{ML}^2\text{T}^{-2}}{\text{AT A}}=[\text{M}^1\text{L}^2\text{T}^{-3}\text{A}^{-2}]$

$\text{C}=\frac{\text{q}}{\text{V}}=\frac{\text{q}}{{\frac{\text{W}}{\text{q}}}}=\frac{\text{q}^2}{\text{W}}=\frac{(\text{AT})^2}{\text{ML}^2\text{T}^{-2}}=[\text{M}^{-1}\text{L}^{-2}\text{T}^{4}\text{A}^{2}]$

$\text{I}=\frac{\text{E}}{\frac{\text{dI}}{\text{dt}}}=\frac{\frac{\text{W}}{\text{q}}}{\frac{\text{dI}}{\text{dt}}}=\frac{\text{ML}^2\text{T}^{-2}\text{T}}{\text{AT A}}=[\text{ML}^2\text{T}^{-2}\text{T}^{-2}]$

Now $\text{ RC}=[\text{M}^1\text{L}^2\text{T}^{-3}\text{T}^{-2}][\text{M}^{-1}\text{L}^{-2}\text{T}^{4}\text{T}^2]=\text{T}^{1}$

And $\frac{\text{L}}{\text{R}}=\frac{[\text{ML}^2\text{T}^{-2}\text{A}^{-2}]}{[\text{M}^1\text{L}^2\text{T}^{-3}\text{A}^{-2}]}=\text{T}^1$

2. 57.3°

Explanation:

We know that,

$\pi\text{ radian}=180^{\circ}$​​​​​​​

$1\text{ radian}=\frac{180}{\pi}=\frac{180}{22}\times7=57.3^{\circ}$

2. $\pm1\%\text{ and }\pm0.1\%$

2. $[\text{M}^{\frac{3}{2}}\text{L}^{\frac{-1}{2}}\text{T}^{-2}],[\text{M}^{\frac{1}{2}}\text{L}^{\frac{-3}{2}}\text{T}^0]$

1. 4.9cm.

Explanation:

Error or absolute error

$|\Delta\text{a}_1|=|5-4.9|=0.1\text{cm},\ |\Delta\text{a}_2|=|5-4.805|=0.195\text{cm}$

$|\Delta\text{a}_3|=|5-5.25|=0.25\text{cm},\ |\Delta\text{a}_4|=|5-5.4|=0.4\text{cm}$

$|\Delta\text{a}_1|$ is minimum. Hence verifies option (a).

1. $\Big(\frac{\text{P}-\text{Q}}{\text{R}}\Big)$

3. $\text{LT}^{-2};\text{L};\text{T}$

Explanation:

As c is added to t, therefore, c has the dimensions of [T]

$\text{As},\frac{\text{b}}{\text{t}}=\text{v}$

$\therefore\text{b}=\text{v}\times\text{t}=\text{LT}^{-1}\times\text{T}=[\text{L}]$

From $\text{v}=\text{at},\text{a}=\frac{\text{v}}{\text{t}}=\frac{\text{LT}^{-1}}{\text{T}}=[\text{LT}^{-2}]$