JEE 2013 :: Advanced 2013 :: Physics 2013

• Advanced 2013 - Physics 2013
• Advanced 2013 - Chemistry 2013
• Advanced 2013 - Mathematics 2013

The mass of a nucleus  $^{A}_{Z}X$ is less than the sum of the mass of (A-Z) number of neutrons and  Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus.A heavy nucleus of mass M can break into two light nuclei of masses m₁ and m₂ only if (m₁+m₂) <M. Also two light nuclei of masses m₃  and m₄ can undergo complete fusion and form a heavy nucleus of mass M ' only if (m₃+m₄)>M'. The masses of some neutral  atoms are given in the table below:

The correct statement is

Answer:

Explanation:

(a)  $_{3}Li^{7}\rightarrow_{2}He^{4}+_{1}H^{3}$

 $\triangle m= [M_{Li}-M_{He}-M_{H^{3}}]$

 =[6.01513-4.002603-3.016050]

  = -1.003523 u

 $\triangle $ m is negative so reaction is not possible

(b)   $_{84}Po^{210}\rightarrow_{83}Bi^{209}+_{1}P^{1}$

 $\triangle$  m is negative so  reaction is not possible

(c)  $_{1}H^{2}\rightarrow_{2}He^{4}+_{3}Li^{6}$

   $\triangle $ m is positive  so reaction is possible

(d)    $_{30}Zn^{70}+_{34}Se^{82}\rightarrow_{64}Gd^{152}$

$\triangle $ m is positive  so reaction is not possible

A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity ω. This can be considered as equivalent to a loop carrying a steady current   $\frac{Q\omega}{2\pi}$. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The applications of the magnetic field induce an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant $\gamma$.

The change in the magnetic dipole  moment associated with the orbit, at the end of the time interval  of the magnetic field change is 

Answer:

Explanation:

$\frac{M}{L}=\frac{Q}{2m}$

$\therefore$   $M=(\frac{Q}{2m})L$

 $\Rightarrow$   $ M\propto L,$   where $ \gamma=\frac{Q}{2m}$

$=\left(\frac{Q}{2m}\right)(l\omega)$

$=\left(\frac{Q}{2m}\right)(mR^{2}\omega)=\frac{Q\omega R^{2}}{2}$

Induced  electric field is opposite.

 Therefore   $\omega'=\omega-\alpha t$

 $\alpha=\frac{\tau}{I}=\frac{(QE)R}{mR^{2}}=\frac{(Q)(\frac{BR}{2})R}{mR^{2}}=\frac{QB}{2m}$

  $\therefore$    $\omega'=\omega-\frac{QB}{2m}.1=\omega-\frac{QB}{2m}$

$M_{f}=\frac{Q\omega' R^{2}}{2}=Q(\omega-\frac{QB}{2m})\frac{R^{2}}{2}$

$\therefore$   $\triangle M=M_{f}-M_{i}=-\frac{Q^{2}BR^{2}}{4m}$

 $M=-\gamma \frac{QBR^{2}}{2}$                      $(as\gamma=\frac{Q}{2m})$

A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity ω. This can be considered as equivalent to a loop carrying a steady current   $\frac{Q\omega}{2\pi}$. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The applications of the magnetic field induce an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant $\gamma$.

The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic  field change is 

Answer:

Explanation:

The induced electric field is given by

 $\oint  E.dl=-\frac{d\phi}{dt}$

or      $E l=-s(\frac{dB}{dt})$

 $\therefore$    $E(2\pi R)=-(\pi R^{2})(B)$

  or             $E= -\frac{BR}{2}$

A thermal powerplant produces electric power of 600kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumer's usage. It can be transported either directly with a cable of large current carrying capacity or using a combination of step up and step down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value.  At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with a power factor unity.All the current and voltage mentioned are rms values.

In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1:10. If the power to the consumers has to be  supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is 

Answer:

Explanation:

During step up

   $\frac{N_{p}}{N_{s}}=\frac{V_{p}}{V_{s}}$

 or   $\frac{1}{10}=\frac{4000}{V_{s}}$

  or V_s= 40000V

 in step down transformer

  $\frac{N_{p}}{N_{s}}=\frac{V_{p}}{V_{s}}=\frac{40000}{200}=\frac{200}{1}$

A thermal powerplant produces electric power of 600kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumer's usage. It can be transported either directly with a cable of large current carrying capacity or using a combination of step up and step down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value.  At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with a power factor unity.All the current and voltage mentioned are rms values.

If the direct transmission method with a cable of resistance 0.4Ω km^-1 is used, the power dissipation (in %)  during transmission is 

Answer:

Explanation:

P=Vi

$\therefore$      $i=\frac{P}{V}=\frac{600\times 10^{3}}{4000}=150A$

 Total resistance  of cables

 $R= 0.4\times20= 8$Ω

 $\therefore$ power loss in cables = i²R

 = (150)²)(8)

  = 180000 W=180 kW

 This loss is 0% of 600 k W

• Advanced 2013 - Physics 2013
• Advanced 2013 - Chemistry 2013
• Advanced 2013 - Mathematics 2013