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An electric lamp is connected to 220 V, 50 Hz supply. Then the peak value of voltage is
Answer : c
Solution : V(0) = Vrms × 2
= 220 × √2
= 310
In a stepup transformer the voltage in the primary is 220 V and the current is 5A. The secondary voltage is found to be 22000V. The current in the secondary (neglect losses) is
A body moves along a circular path of radius 10 m and the coefficient of friction is 0.5. What should be the angular velocity of the body, if it is not to slip on the surface?
Higher the voltage, higher is the KE. Higher the work function, smaller is the KE.
A semi circle arc of radius 'a' is charged uniformly and the charge per unit length is λ. The electric field at its centre is
A gun fires the bullets each of mass 10 g with a velocity of 50 m−s^{1}. If 1000 bullets are fired per second, then thrust on the shoulder will be
Mass of bullet
m=0.01Kg
v=50m/s
n=1000
Force on the man is equal to the change in momentum of the bullets coming out of the gun
F = mvn/t
F=(0.01×50×1000)/1
Therefore, F = 500N
An electric dipole is placed along the Xaxis at the origin O. A point P is at a distance of 20 cm from this origin such that OP makes an angle π/3 with the X axis. If electric field at P makes an angle θ with Xaxis, the value of θ is
A metre stick AB hinged (without friction) at the end A as shown in the figure is kept horizontal by means of a string tied to the end B, the other end of the string being tied to a hook. The string is carefully cut (or burnt), and the scale executes angular oscillations about an axis passing through the end A. What is the speed of the end B when the metre stick assumes vertical position immediately after the string is burnt? (Acceleration due to gravity = 10 ms^{–2})
The centre of gravity of the metre stick is at its middle. When the string is burnt, the height of the centre of gravity is reduced by 0.5 m, thereby reducing the gravitational potential energy of the metre stick by Mgh= M×10×0.5 = 5M joule, where M is the mass of the metre stick.
The decrease in the potential energy is equal to the increase in kinetic energy. Therefore in the vertical position of the metre stick, its kinetic energy
(½)Iω^{2} = 5M
Since the moment of inertia of a uniform bar of length L about a normal axis through ite mid point is ML^{2}/12, its moment of inertia about a parallel axis through its end is ML^{2}/3. [You will get this by applying the parallel axis theorem (see the post dated 2o^{th} January 2008)]
Therefore, we have (½)(ML^{2}/3)ω^{2} = 5M
Since ω = v/L and L = 1 metre, the above equation becomes
v^{2}/6 = 5, from which v = √30 = 5.4 ms^{–1} (approximately).
The symbolic representation of four logic gates are given below:
The logic symbols for OR, NOT and NAND gates are respectively:
In a capillary tube, water rises upto 3 mm. The height of water that will rise in another capillary tube having onethird radius of the first is
The narrower the tube, the higher is the rise in water.
When a glass capillary
h=2Tcosθ/Rpg
h∝1/R
h2/h1=R1/R2
3/h2=1/3
h2=9mm
If a long hollow copper pipe carries a current then magnetic field produced will be
A chain reaction in fission of uranium is possible because
A nuclear chain reaction occurs when one single nuclear reaction causes an average of one or more subsequent nuclear reactions, thus leading to the possibility of a selfpropagating series of these reactions.
Two bodies of mass 10 kg and 5 kg moving in concentric orbits of radii R and r such that their periods are the same. Then the ratio between their centripetal acceleration is
When there is an electric current through a conducting wire along its length then an electric field must exist
As current is flowing through it , it means charges are flowing along its length therefore heir must be some electric field parallel to the length of the wire. Hence correct option is B.
In the following question, a Statement of Assertion (A) is given followed by a corresponding Reason (R) just below it. Read the Statements carefully and mark the correct answer
Assertion(A):All radioactive elements are finally converted into lead.
Reason(R):All the elements above lead are unstable
For all 4n,4n+2 series elements, on emission of different kind of rays, they form different isotopes of lead and also all element above lead are unstable .
But, 4n+1 series elements (Neptunium series), end in Bi (Bismuth not lead)
The moment of inertia of a uniform disc about an axis passing through its centre and perpendicular to its plane is 1kg  m^{2}. It is rotating with an angular velocity 100 radians/second. Another identical disc is gently placed on it so that their centres coincide. Now these two discs together continue to rotate about the same axis. Then the loss in kinetic energy in kilo joules is
When a gas expands adiabatically, it loses internal energy equal to the work done by the gas. In this case the temperature of the gas decreases.
When a gas is compressed adiabatically, it gains internal energy equal to the work done on the gas. In this case the temperature of the gas increases.
If T is the reverberation time of an auditorium of volume V, then
Generally reverberation time is proportional to linear dimensions of the room. Hence, it will be proportional to the volume of the room.
Consider the following statements.
(i) All isotopes of an element have the same number of neutrons.
(ii) Only one isotope of an element can be stable and nonradioactive.
(iii) All elements have isotopes.
(iv) All isotopes of Carbon can form chemical compounds with Oxygen16.
The correct option regarding an isotope is?
All elements have isotopesatoms of the same element can have different number of neutrons.
Carbon form compounds (carbon monoxide and carbon dioxide) with oxygen. This means that all isotopes of carbon can form compounds (carbon monoxide and carbon dioxide) with all isotopes of oxygen, including oxygen16.
The orbital velocity of an artificial satellite in a circular orbit just above the earth’s surface is v_{0}. The orbital velocity of satellite orbiting at an altitude of double the radius of earth is V_{0}/√n Then the value of n is :
A 300 W carrier is modulated to a depth 75%. The total power in the modulated wave is:
A conducting rod of 1 m length and 1 kg mass is suspended by two vertical wires through its ends. An external magnetic field of 2T is applied normal to the rod. Now, the current to be passed through the rod so as to make the tension in the wires zero is : (Take g = 10 ms^{–2}) :
Magnetic force on rod = BIℓ
Weight of the rod = mg
For no tension in wire, BIℓ = mg
At what speed a ball must be projected vertically upward so that distance travelled by it in 5^{th} second is equal to distance travelled in sixth second (in m/s):
time of flight = (5 sec) × 2 = 10 sec
µ = gt = 9.8 × 5 = 49 m/s
In thermodynamic process pressure of a fixed mass of gas is changed in such a manner that the gas releases 30 joule of heat and 18 joule of work was done on the gas. If the initial internal energy of the gas was 60 joule, then, the final internal energy (in J) will be:
ΔQ = –30J
Δw = –18
U_{i} = 60 J
U_{f} = ?
ΔQ = Δw + U_{f} – U_{i}
–30 = –18 + U_{f} – 60 ⇒ U_{f} = 48 J
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