Units and Measurement

Magnetic energy per unit volume $=\frac{{\mathrm{B}}^2}{2{?}_0}$

$Torque\overrightarrow{\tau }=\overrightarrow{r}*\overrightarrow{F}$

$\left[\tau \right]=\left[L\right]\left[ML{T}^{-2}\right]$

$\left[\tau \right]=\left[M{L}^2{T}^{-2}\right]$

as, $\tau =m{\mathcal{l}}^2{t}^{-2}$

$\frac{d\tau }{\tau }*100=\left|\frac{dm}{m}*100\right|+2\left|\frac{d\mathcal{l}}{\mathcal{l}}*100\right|+2\left|\frac{dt}{t}*100\right|$

= 5 + 2 * 5 + 2 * 5

= 25 %

A. Torque   $\underset{}{\overset{}{\to }}$   (iii) Nm

B. Stress  $\underset{}{\overset{}{\to }}$ (iv) Nm^-2

C. Latent Heat  $\underset{}{\overset{}{\to }}$ (ii) J kg^-1

D. Power   $\underset{}{\overset{}{\to }}$ (i) Nm S^-1

Velocity gradient = $\frac{dv}{dx}=\frac{L{T}^{-1}}{L}=T^{-1}$

Decay constant $\left(\lambda \right)=\frac{0.693}{T_{\frac{1}{2}}}=T^{-1}$

Least count of Vernier = 0.1mm

$\Rightarrow$ Reading of Vernier Scale = 5 * 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 * 102 cm.

NLM² for (1)                                   

mg – T = ma

Þ 2 * 10 – T = 2a

Þ T + 2a = 20                   - (1)

Rotation equestion for (2)

T * R = I_cma

Þ T R =  $I_{CM}\frac{a}{R}$

  $\left[\alpha =\frac{a}{R},Noslippingcondition\right]$

T = $\frac{Ma}{2}$  

$T=\frac{4}{2}a$

T = 2a                  - (2)

from (1) & (2)

2T = 20

T = 10N

$\frac{sin{\theta }_C}{V_B}=\frac{sin90°}{V_A}\Rightarrow sin{\theta }_C=\frac{V_B}{V_A}=\frac{1.5*10^{10}}{2.0*10^{10}}=\frac{3}{4}$

According to question, we can write

$\theta >\theta =si{n}^{-1}\left(\frac{3}{4}\right)$

Let, L₁ & L₂ one distance for system (1) & (2) respectively

T₁ & T₂ are time for system (1) & (2) respectively

Given : v2 = $\frac{n}{m^2}{v}_1$

$\Rightarrow \left(\frac{L_2}{T_2}\right)=\frac{n}{m_2}\left(\frac{4}{T_1}\right)$ $\left[?v_2=\frac{L_2}{T_2},v_1=\frac{L_1}{T_1}\right]$

$\Rightarrow \frac{L_2}{L_1}=\left(\frac{n}{m_2}\right)\left[\frac{T_2}{T_1}\right]--\left(i\right)$

And a2 = $\frac{a_1}{mn}\Rightarrow \frac{v_2}{T_2}=\frac{v_1}{T_{1}mn}$

$\Rightarrow \frac{T_1}{T_2}=\frac{v_1}{v_2}\left(\frac{1}{mn}\right)$ $\left[?\frac{v_1}{v_2}=\frac{m^2}{n}\right]$

$\Rightarrow \frac{n^2}{m}{T}_1=T_2$

from (1)

$\frac{L^2}{L_1}=\frac{n}{m^2}\left[\frac{n^2}{m}\right]\left[?\frac{T_2}{T_1}=\frac{n^2}{m}\right]$

$\Rightarrow \frac{n^3}{m^3}{L}_1=L_2$

 $h=50t-\frac{1}{2}\text{?}*10t^2=80m$

Now velocity B with respect to A

$V_{BA}=V_B-V_A$

= 50 – 30

= 20 m/sec

Now $A_{BA}=a_B-a_A$

= $10\left(-\widehat{j}\right)-\left(-10\widehat{j}\right)$

= 0

Time taken to collide.

$h=V_{BA}t$

80 = 20t

t = 4 sec

Total time = 4 + 2

= 6 sec

Least count = $\frac{0.5}{50}mm$ = 0.01 mm

Diameter, d = 1.5 + 7 × 0.01

= 1.57

$\therefore$ Surface Area = (2r) $\mathcal{l}$

= 3.4 cm²