Physics Test - 68

Given, \(A_{1}=56, A_{2}=238\)

We know,

\(\frac{R_{1}}{R_{2}}=\left(\frac{A_{1}}{A_{2}}\right)^\frac{1}{3}\)

\(\Rightarrow\left(\frac{56}{238}\right)^\frac{1}{3}=(0.235)^\frac{1}{3}=0.617\)

As nuclear density is same for all nuclei, therefore, \(\frac{\rho_{1}}{\rho_{2}}=1\)

Given,

No. of electrons \(= 25 \times 10^{31}\)

We know that:

Charge of one electron \(=1.6 \times 10^{-19}\) Coulomb.

Therefore,

Charge of \(25 \times 10^{31}\) electrons will be:

\(Q=\) No. of electrons\( \times\) Charge of one electron

\(Q= 25 \times 10^{31} \times 1.6 \times 10^{-19} \)

\(Q=40 \times 10^{12} C\)

Given,

Length \({l}=16.2 \) cm

Breadth \(b=10.1 \) cm

We can write with least count,

\(l=16.2 \pm 0.1 \) cm

Now, percentage uncertainty in length\(=\frac{0.1}{16.2} \times 100\)

\(=0.617\approx 0.6 \%\)

Similarly, we can write with least count,

\(b=10.1 \pm 0.1 \) cm

The percentage uncertainty in breadth\(\frac{0.1}{10.1} \times 100\)

\(=0.990099\approx 1 \%\)

Now we can write,

\(l=16.2\) cm \( \pm 0.6 \% \)

\(b=10.1\) cm \( \pm 1 \% \)

Area of a rectangle \(=l \times b\)

\(=16.2\pm 0.6 \% \times 10.1 \pm 1 \% \)

\(=163.62 \) cm\(^{2}\) \(\pm 1.6 \%\)

Thus,the percentage uncertainty in the area of this rectangle \( =1.6 \%\)

Work \(=\) Loss in \(K . E .=\frac{1}{2} I \omega^{2}\)

\(I_{A}=\frac{2}{5} M R^{2}=0.4 M R^{2}\)

\(I_{B}=\frac{1}{2} M R^{2}=0.5 M R^{2}\)

\(I_{C}=M R^{2}\)

\(I_{C}>I_{B}>I_{A}\)

\(W_{C}>W_{B}>W_{A}\)

Given,

Linear charge density, \(\lambda=8 \mathrm{nC} / \mathrm{m}=8 \times 10^{-9} \mathrm{C} / \mathrm{m}\)

The relation between surface charge density and linear charge density can be given as

\(\frac{2 k \lambda}{r}=\frac{\sigma}{\varepsilon_0}\)........(i)

where,

\(k=\) Coulomb's constant,

\(\lambda=\) linear charge density,

\(\sigma=\) surface charge density,

\(\varepsilon_0=\) absolute electrical permittivity of free space

and \(r=\) distance.

Substituting the values in Eq. (i), we get

\(\sigma  =\frac{2 k \lambda \varepsilon_0}{r} \)

\(=\frac{2 \times 9 \times 10^9 \times 8 \times 10^{-9} \times 8.85 \times 10^{-12}}{3} \)

\(=0.424 \times 10^{-9} \mathrm{Cm}^{-2}=0.424 \mathrm{nCm}^{-2}\)

\(\left(\Delta I_{C u}=(\Delta)_{A l}\right.\)

\(I_{C u} a_{C u} \Delta T=I_{A I} a_{A I} \Delta T\)

\(88 \times 1.7 \times 10^{-5}=I_{2} \times 2.2 \times 10^{-5}\)

\(I_{A l}=68 \mathrm{~cm}\)

Since the initial velocity of the car is zero, its velocity at the end of the first time interval \(t_1\) is \(v =0+ a_ 1 t_1= a_1 t_1\). This is the initial velocity for the next time interval \(t_2\), since the final velocity is zero, we have, from \(v = u + at\)

\(0=a_{1} t_{1}-a_{2} t_{2} \quad\left(\because u=a_{1} t_{1}\right)\)

Now, the distance covered in the first time interval \(t_1\) is given 

\(2 a_{1} s_{1}=v^{2}-u^{2}=a_{1}^{2} t_{1}^{2}\)

\(\\therefore v=a_{1} t_{1}\) and u=0

\(s_{1}=\frac{1}{2} a_{1} t_{1}^{2}\)....(i)

The distance covered in the next time interval \(t_{2}\) is given by \(-2 a_{1} s_{2}=0-a_{1}^{2} r_{1}^{2}\)

\(s_{2}=\frac{1}{2} \frac{a_{1}^{2}}{a_{2}} t_{1}^{2}=\frac{1}{2} \frac{a_{2}^{2} t_{2}^{2}}{a_{2}}\)     \(\left(\because a_{1} t_{1}=a_{2} t_{2}\right)\)

or \(s_{2}=\frac{1}{2} a_{2} t_{2}^{2}\)

From (i) and (ii) we get

\(\frac{s_{1}}{s_{2}}=\frac{a_{1} t_{1}^{2}}{a_{2} t_{2}^{2}}\).... (iii)

Multiplying \(a_{2} a_{1}\) in equation (iii) with both the numerator and denominator, we get

\(\frac{s_{1}}{s_{2}}=\frac{a_{2} a_{1}}{a_{2} a_{1}} \frac{a_{1} t_{1}^{2}}{a_{2} t_{2}^{2}}=\frac{a_{2}}{a_{1}} \frac{a_{1}^{2} t_{1}^{2}}{a_{2}^{2} t_{2}^{2}}=\frac{a_{2}}{a_{1}}\left(\because a_{1} t_{1}=a_{2} t_{2}\right)\)

Thus, we have \(\frac{s_{2}}{s_{1}}=\frac{a_{1}}{a_{2}}=\frac{t_{2}}{t_{1}}\)

For block \(\mathrm B\) to move slowly \(\mathrm {T=f_1=250 \mathrm{~N}}\)
Maximum possible friction on shoes of the man \(\mathrm{f_{2 \max }=0.6 \times 500=300 \mathrm{~N}}\)
For man to remain at rest relative to the ground \(\mathrm{f_2=T=250 \mathrm{~N}}\)
Acceleration of platform \(\mathrm A\) is, \(30\times\mathrm{ ~a=f_2-f_3}\)\(=250-0.2 \times 800\)
\(\mathrm {a}=3 \mathrm{~m} / \mathrm{s}^2\)
Man will fall off the platform once he moves \(5 \mathrm m\). Hence, time required is given by \(\mathrm {\frac{1}{2} a t^2=5} \) \(\Rightarrow\frac{1}{2} \times 3 \times\mathrm t^2=5\) \(\Rightarrow \mathrm {t=\sqrt{\frac{10}{3}}} s\)

From the ideal gas law, the universal gas constant \(R=\frac{P V}{n T}\)

Unit of Pressure \((P)=\) \(\mathrm{Nm}^{-2}\)

Unit of Temperature \((T)=\)  \(\mathrm{K}\) 

Unit of Volume \(({V})=\)  \(\mathrm{m}^3\)

Unit of Number of moles \(({n})=\)  \(\mathrm{mol}\)

Substituting the units in (1),

\( R=\frac{\left(\mathrm{Nm}^{-2}\right)\left(\mathrm{m}^3\right)}{(\mathrm{mol})(\mathrm{K})}=\mathrm{Nm} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\)

So, \(\mathrm{NmK}^{-1} \mathrm{~mol}^{-1}\) is the unit of \({R}\).

\(\mathrm{Nm}\) is also the unit of energy and can be denoted by joules \((J)\). Thus, \(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) is also a unit of \({R}\).

An erg is a unit of energy equal to \(10^{-7}\) joules. So, \(\mathrm{{erg} K^{-1} \mathrm{~mol}^{-1}}\) is a unit of \({R}\). Thus, all the given options are correct.

We know that:

A capacitor consists of two plates of a conductor and a dielectric insulator between them:

\(C=\frac{\epsilon A}{d}\)

Where

\(C=\) capacitance in farad

\(\epsilon=\) Permittivity of dielectric

\(A=\) area of plate overlap in square meters

\(d=\) distance between plates in meters

If the area of the plates is halved i.e., \(\frac{A}{2}\) and the distance between them is halved i.e., \(\frac{d}{2}\), then new capacitance is:

\(C^{\prime}=\frac{\epsilon \frac{A}{2}}{\frac{d}{2}} \)

\(\therefore C^{\prime}=C\)