Physics Class 12 Mini Mock - 5 - Free MCQ Practice Test with solutions,

P=I_rms V_rms cos ϕ
Where, ϕ=90
cosϕ=0
therefore,P=0
hence the correct answer is option A.

and as we know 1Å = 0.1 nm, so 1.227Å = 0.122 nm

Infrared: Night vision goggles pick up the infrared light emitted by our skin and objects with heat. In space, infrared light helps us map the dust between stars.
Visible: Our eyes detect visible light. Fireflies, light bulbs, and stars all emit visible light.

Ultraviolet: Ultraviolet radiation is emitted by the Sun and are the reason skin tans and burns. "Hot" objects in space emit UV radiation as well.

X-ray: A dentist uses X-rays to image your teeth, and airport security uses them to see through your bag. Hot gases in the Universe also emit X-rays.

Gamma ray: Doctors use gamma-ray imaging to see inside your body. The biggest gamma-ray generator of all is the Universe.

The time period of the charged particle is
given by  

Thus, time period is independent of both v and R.

The nuclide (nucleus) consists of neutrons and protons (when combined called nucleons).
Thus,
No. of protons in 92U²³⁸ = 92,
No. of neutrons = 146 (238 – 92)
No. of nucleons = 238 (146 + 92)

Power of Cornea, P₁​=40D
and power of the eye lens, P₂​=20D
P_eff​=P₁​+P₂
⇒Peff​=40D+20D=60D
So, f= ​100/ P_eff ​=100​/60=1.67cm
 

A real image is defined as one that is formed when rays of light are directed in a fixed point.  A real image can be projected or seen on a screen. The best example of a real image is the one formed on a cinema screen.
A virtual image is defined as the opposite of a real image, therefore an image that cannot be obtained on a screen is referred to as a virtual image. The explanation for this is the fact that the rays of light that form a virtual image never converge therefore a virtual image can never be projected onto a screen. The best example of a virtual image is your reflection in the mirror.
Real images are produced by intersecting rays while virtual images are produced by diverging rays.
Real images can be projected on a screen while virtual ones cannot.
Real images are formed by two opposite lenses, concave and convex.
Virtual images are always upright while real images are always inverted.
 

The standard convention states that the electric potential at a point infinitely far from a charge is considered to be:

• Zero

This means that as you move further away from a charge, the potential energy associated with that charge diminishes to zero.

In physics and in electronic engineering, dark current is the relatively small electric current that flows through photosensitive devices such as a photomultiplier tube, photodiode, or charge-coupled device even when no photons are entering the device; it consists of the charges generated in the detector when no outside radiation is entering the detector.

It is referred to as reverse bias leakage current in non-optical devices and is present in all diodes. Physically, dark current is due to the random generation of electrons and holes within the depletion region of the device.

In this case, R₁ and R₂ are unchanged
So, f will remain unchanged for both pieces
of the lens
∴ f = f '

This is combination of two lenses of equal
focal lengths

Current in the circuit

P.D. across first cell = E – ir₁

In Optics, the angle of incidence to which the angle of refraction is 90° is called the critical angle. The ratio of velocities of a light ray in the air to the given medium is a refractive index. Thus, the relation between the critical angle and refractive index can be established as the Critical angle is inversely proportional to the refractive index.
Critical Angle and Refractive Index
The relationship between critical angle and refractive index can be mathematically written as –
SinC=1/μ_a^b
μ_a^b =1/SinC
Where,
C is the critical angle.
μ is the refractive index of the medium.
a and b represent two mediums in which light ray travels.
 

Refractive index of glass with respect to liquid:
n = 1.52/1.63 = 0.9325
thus, critical angle, i = sin^-1(n) = 68.8 degrees

The incident ray and emergent ray are parallel to each other but latteray displaced due to reflaction at two surfaces. So, the angle between them is Zero.

A thick Plano convex lens made of crown glass (refractive index 1.5) has a thickness of 3cm at its centre. The radius of curvature of its curved face is 5cm. An ink mark made at the centre of its plane face, when viewed normally through the curved face, appears to be at a distance ‘x’ from the curved face, and we have to determine the value of ' x'
According to the picture, the ray of light gets refracted at the interface between the air and the lens from the object 'p' and 'I' is the refracted image of 'p'
object distance 'u'= BO
Image distance 'v'/'x' = BI
we know
n2/v - n1/u = (n2- n1) / R
or 1/v - 1.5/(-3 )= (1-1.5)/ (-5) [where n2= 1, n1= 1.5 ,u= -3 ,R= -5]
so, 1/v = -6/15
or, v = -2.5
so, x is equal to 2.5 cm

Lenz’s law is used for determining the direction of induced current.
Lenz’s law of electromagnetic induction states that the direction of induced current in a given magnetic field is such that it opposes the induced change by changing the magnetic field.
Following is the formula of Lenz’s law:
ϵ=−N (∂ϕ_B/∂t)
Where,

• ε is the induced emf
• ∂ΦB is the change in magnetic flux
• N is the number of turns in the coil
   

Lenz’s law finds application in electromagnetic braking and in electric generators

As I = dQ/dt ;
dQ − Idt ;
dQ = (2t² − 3t + 1)dt

Q.= 
= [2/5(125 − 27) − 32(25 − 9) + 2]
= 43.34C

The ratio of angular dispersion to the angle of deviation for the mean wavelength is called Dispersive Power. Represented by 

where δ_v − δ_r is the angular dispersion for violet and red lights and δ_r the deviation suffered by the mean light.

Consider refraction of light from infinity at first surface,
(μ2/v)​​−( μ1/u​)​=(μ2​−μ1/R)​​
(1.5​/ v1​)-(1.75/−∞)​=1.5−1.75​/−R
Consider refraction of this image from second surface,
​(1.75​/ v2)−(​1.5/v1)​=(1.75−1.5​)/R
Hence, v2​=3.5R
This is the image of light coming from infinity, therefore the focal length.
It is converging since focal length is positive.
 

In a parallel plate capacitor the capacity of capacitor.

The capacity of capacitor increases if area of the plate is increased.

As the electron is moving along the direction of the magnetic field, it will experience no magnetic force, but due to an electric force acting on it opposite to the direction of electric field (as it is a negatively charged particle) the velocity of the electron will decreases.

λ for U.V is less than λ for visible light 
ν for U.V is greater than ν for visible light 
∴ potential is greater for U.V light.
as K.Eα1​/λ

Beta(central Maxima) = 2 * lambda * D(distance of the screen) / d(distance between slits)
beta = 2 m, wavelength = 500nm = 5 ×10^-7m & d = 0.1mm = 1×10^-4m
► 2 = 2 × 5 × 10^-7 * D/10^-4
► 1 = 5 × 10^-3 D 
► D = 1000/5
► D = 200m

The dipole moment is defined as the product of a charge and distance. The dimension of charge (current*time) is [I T] and the dimension of distance is [L]. Therefore the dimension of dipole moment is [L T I]. Its unit in the CGS and the SI system are esu*cm and C*m respectively.

When he sent alpha particles during his gold foil experiment he observed that most of the rays moved away from the nucleus.As we know that like charges always repel ,he concluded that the charge the nuclei is same as the charge of alpha particles

An electroscope is a device that measures the potential difference.
If it is connected in parallel to the capacitor, the potential across it will be equal to the potential across the capacitor, which is equal to the potential across the battery.
On decreasing the battery potential, the potential difference across the electroscope reduces, and hence the reading reduces. 

Electric potential is a scalar quantity and its value is solely dependent on the charge near it and the distance from that charge. In this case, the point is equidistant from the two point charges and the point charges have the same value but opposite nature. Therefore equal but opposite potentials are generated due to the charges and hence the net potential at midpoint becomes zero.