Time taken for 360° shift = 24 h
Time taken for 1° shift = 24/360 h = 4 min.

(i) As the lens forms a real iamge, it must be a convex lens.
(ii) From the graph, when u = 20 cm , we have v = 20 cm.
For the convex lens forming a real iamge, u is negative and v and f are positive.
U = -20 cm v = +20cm
Using this lens formula,
1/f = 1/v – 1/u = 1/20 – 1/-20 = 1/10 or f = + 10 cm

doubled

Consider the diagram.
Bandwidth = \(\omega _{ 2 }-{ \omega }_{ 1 }\),

Where , \(\omega_1\) and  \(\omega_2\)  correspond to frequencies at which magnitude of  current is \(\frac { 1 }{ \sqrt { 2 } } \) times of maximum value.
\({ I }_{ rms }=\frac { { I }_{ max } }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \approx 0.7A\)
Clearly, from the diagram, the corresponding frequencies are 0.8 rad/s and rad/s.
\(\triangle \omega \) = B and width = 1.2 - 0.8 = 0.4 rad/s

As, the value of R is increased, the current flowing in the circuit will decrease. And the potential gradient, i.e. potential drop per unit length also decreases, so that the balance length will increase. Thus, J will shift towards B.

When an electron falls from a higher energy to a lower energy level, difference in energies appears in the form of electromagnetic radiation only. This is because electrons (being charged) interact only electromagnetically.

About \(10 kΩ.\) This resistance is mainly due to skin through which current enters and leaves our body.

(a) As explained in theory, \(\alpha \)-decay is caused by the quantum mechanical tunnelling of an alpha particle through a repulsive Coulomb barrier. The rate of tunnelling would depend upon the height and width of the barrier. The decay cannot be all at once. And that is the reason why half life of \(_{ 92 }{ { U }^{ 238 } }\) against \(\alpha \)-decay is large.
(b) The possible nuclear reaction is \(_{ 92 }{ { U }^{ 238 } }\rightarrow _{ 92 }{ { U }^{ 237 } }+_{ 0 }{ { n }^{ 1 } }\)
The data shows that \(m\left( _{ 92 }{ { U }^{ 237 } } \right) +m\left( n \right) \) is greater than \(m\left( _{ 92 }{ { U }^{ 238 } } \right) .\) Therefore, the decay is not allowed energetically. Rather, some external energy has to be supplied to separate a neutron from \(_{ 92 }{ { U }^{ 238 } }\).

Let m₁ = m₂ = m, r= 3 mm = 3 \(\times\) 10^-3 m
For hanging in air in balanced position,
F= 50 gf = 50 \(\times\) 10^-3\(\times\)9.8 N
\(As\quad F=\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } \\ \therefore \quad 50\times { 10 }^{ -3 }\times 9.8={ 10 }^{ -7 }\frac { m.m }{ { \left( 3\times { 10 }^{ -3 } \right) }^{ 2 } } \) 
m²=9\(\times\)59.8\(\times\)10^-1 , m = 6.64 Am.

(a) Refractive index of the glass fibre, μ₁ = 1.68
Refractive index of the outer covering of the pipe,  μ₂ = 1.44
Angle of incidence = i
Angle of refraction = r
Angle of incidence at the interface = i’
The refractive index (μ) of the inner core − outer core interface is given a 
\(\mu =\frac { { \mu }_{ 1 } }{ { \mu }_{ 2 } } =\frac { 1 }{ sin \ i } \)
sin i' = \(\frac { { \mu }_{ 1 } }{ { \mu }_{ 2 } } \)
= \(\frac{1.44}{1.68}\) = 0.8571
∴ i' = 59^o 
For the critical angle, total internal reflection (TIR) takes place only wheni > i', i.e., i > 59°
Maximum angle of reflection, r_max = 90^o - i' = 90^o - 59^o = 31^o
Let, i_max be the maximum angle of incidence.
The refractive index at the air - glass interface, μ₁ = 1.68
We have the relation for the maximum angles of incidence and reflection as:
\({ \mu }_1\) = \(\frac { sin{ i }_{ max } }{ sin{ r }_{ max } } \)
sin i_msx = μ₁ sin r_max
= 1.68 sin 31^o
= 1.68 x 0.5150
= 0.8652
∴ i_max = sin^-1 0.8652 ≈ 60^o
Thus, all the rays incident at angles lying in the range 0 < i < 60° will suffer total internal reflection.
(b) If the outer covering of the pipe is not present, then:
Refractive index of the outer pipe, μ₁ = Refractive index of air = 1
For the angle of incidence i = 90°, we can write Snell’s law at the air − pipe interface as:
\(\frac { sin \ i }{ sin \ r } ={ \mu }_{ 2 }\) = 1.68
\(sinr=\frac { { sin90 }^{ o } }{ 1.68 } =\frac { 1 }{ 1.68 } \)
r = sin^-1 (0.5952)
= 36.5^o
∴ i' = 90^o - 36.5^o = 53.5^o
Since i' > r, all incident rays will suffer total internal reflection.

The focal length f = -15/2 cm = -7.5 cm
(i) The object distance u = -10 cm. Then Eq gives
\(\frac { 1 }{ v } +\frac { 1 }{ 10 } =\frac { 1 }{ -7.5 } \)
or \(v=\frac { 10\times 7.5 }{ -2.5 } =-30\) cm
The image is 30 cm from the mirror on the same side as the object
Also, magnification m = \(\frac { v }{ u } =-\frac { (-30) }{ (-10) } =-3\)
The image is magnified, real and inverted.
(ii) The object distance u = -5 cm. Then from Eq
\(\frac { 1 }{ v } +\frac { 1 }{ -5 } =\frac { 1 }{ -7.5 } \)
or \(v=\frac { 5\times 7.5 }{ (7.5-5) } =15\) cm
This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification m = 15 \(-\frac { v }{ u } =-\frac { 15 }{ (-5) } =3\)
The image is magnified, virtual and erect.

Working Let initially positively charged is accelerated towards D₂ and enter into it.
Now, the charged particle experiences magnetic Lorentz force due to a strong normal magnetic field. It performs circular motion. The time taken by the charge particle to complete half revolution is equal to half of time period of AC oscillator between two dees.
The charge d particle again accelerated towards D₁ as D₂ acquires positive and d negative polarity. Thus, the charge particle is brought again and again in the small region of oscillating electrical field by strong normal magnetic field.
The charged particle repeatedly passes through oscillating electrical field. It traversed on spiral path and finally having radius of its circular path becomes equal to the radius of dees and finally comes out through window W and strikes to the target.
∵ Maximum KE of charged particle
= \(\frac{q^{2}B^{2}r_{0}^{2}}{2m}\)
where, r₀ = radius of dees
∴ Radius of dees is limited, therefore, KE_max also have limited value.

The first refracting ABC forms the image I₁ of the object O. The image I₁ acts as virtual object for the second refracting surface ADC, which forms the real image I as shown in the diagram
For refraction at ABC 
\(\frac { { n }_{ 2 } }{ { v }_{ 1 } } -\frac { { n }_{ 1 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 1 } } \)
For refraction at ADC
\(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ { v }_{ 1 } } =\frac { { n }_{ 1 }-{ n }_{ 2 } }{ { R }_{ 2 } } \)
Adding equation (i) and equation (ii)
 \(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
We know, If  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

\(Here,\ { i }_{ p }=60°,\delta m=? \ A=60°\)
\( \mu =tan{ i }_{ p }=tan60°=\sqrt { 3 }\)
\( For \ prism \ formula,\ \mu =\frac { sin(A+\delta m)/2 }{ sin \ A/2 }\)
\(\sqrt { 3 } =\frac { sin(60°+\delta m)/2 }{ sin30° }\)
\(\frac { sin(60°+\delta m) }{ 2 } =\sqrt { 3 } \times \frac { 1 }{ 2 } =sin60°\)
\(\frac { 60°+\delta m }{ 2 } =60°\ \delta m=2\times 60°-60°=60°\)