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Filename: competition/left-sidebar.php
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Backtrace:
File: /var/www/html/public_html/application/views/competition/left-sidebar.php
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Function: _error_handler
File: /var/www/html/public_html/application/views/competition/entrance_exam_questions.php
Line: 6
Function: view
File: /var/www/html/public_html/application/controllers/Study.php
Line: 6544
Function: view
File: /var/www/html/public_html/index.php
Line: 315
Function: require_once
Multiple Choice QuestionsSeverity: Notice
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Filename: pages/questions.php
Line Number: 82
Backtrace:
File: /var/www/html/public_html/application/views/pages/questions.php
Line: 82
Function: _error_handler
File: /var/www/html/public_html/application/views/competition/entrance_exam_questions.php
Line: 28
Function: view
File: /var/www/html/public_html/application/controllers/Study.php
Line: 6544
Function: view
File: /var/www/html/public_html/index.php
Line: 315
Function: require_once
The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 µT. What will be its value at the centre of the loop?
250 µT
150 µT
125 µT
125 µT
A.
250 µT
A charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to a magnetic field B. The time taken by the particle to complete one revolution is
D.
In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be
200Ω
100Ω
500 Ω
500 Ω
B.
100Ω
The potential drop across the resistance R is 2V. Now, no current flows through the galvanometer .and if V1 is the potential drop across the resistor R and V is the total potential then
A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be
nB
n2B
2nB
2nB
B.
n2B
The magnetic field at the centre of circular coil is
When wire of length i bents into a circular loops of n turns, then
l = n × 2 π r
⇒ r = 1/ n x 2 π
Thus, new magnetic field
Severity: Notice
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Filename: pages/questions.php
Line Number: 356
Backtrace:
File: /var/www/html/public_html/application/views/pages/questions.php
Line: 356
Function: _error_handler
File: /var/www/html/public_html/application/views/competition/entrance_exam_questions.php
Line: 28
Function: view
File: /var/www/html/public_html/application/controllers/Study.php
Line: 6544
Function: view
File: /var/www/html/public_html/index.php
Line: 315
Function: require_once
One conducting U tube can slide inside another as shown in figure, maintaining electrical contacts between the tubes. The magnetic field B is perpendicular to the plane of the figure. if each tube moves towards the other at a constant speed V, then the emf induced in the circuit in terms of B,
B
-B
zero
zero
D.
zero
A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms needed to be connected in series with the coil will be
103
105
99995
99995
D.
99995
A current I ampere flows along an infinitely long straight thin-walled tube, then the magnetic induction at any point inside the tube is
infinite
zero
B.
zero
Let R be the radius of a long thin cylindrical shell. To calculate the magnetic induction at a distance r (r < R) from the axis of cylinder, a circular shell of radius r is shown:
Since no current is enclosed in the circle so, from Ampere's circuital law, magnetic induction is zero at every point of the circle. Hence, the magnetic induction at any point inside the infinitely long straight thin-walled tube (cylindrical) is zero.
A charged oil drop is suspended in a uniform field of 3 × 104 V/m so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge = 9.9 × 10−15 kg and g = 10 m/s2 )
3.3 × 10−18 C
3.2 × 10−18 C
1.6 × 10−18 C
1.6 × 10−18 C
A.
3.3 × 10−18 C
In steady state, electric force on drop = weight of drop
∴ qE = mg
Severity: Notice
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Filename: pages/questions.php
Line Number: 356
Backtrace:
File: /var/www/html/public_html/application/views/pages/questions.php
Line: 356
Function: _error_handler
File: /var/www/html/public_html/application/views/competition/entrance_exam_questions.php
Line: 28
Function: view
File: /var/www/html/public_html/application/controllers/Study.php
Line: 6544
Function: view
File: /var/www/html/public_html/index.php
Line: 315
Function: require_once
A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity then
its velocity will decrease.
its velocity will increase.
it will turn towards right of direction of motion.
it will turn towards right of direction of motion.
A.
its velocity will decrease.
In a uniform magnetic field of induction B a wire in the form of semicircle of radius r rotates about the diameter of the circle with angular frequency ω. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R the mean power generated per period of rotation is
B.
Severity: Notice
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Filename: competition/right-sidebar.php
Line Number: 7
Backtrace:
File: /var/www/html/public_html/application/views/competition/right-sidebar.php
Line: 7
Function: _error_handler
File: /var/www/html/public_html/application/views/competition/entrance_exam_questions.php
Line: 46
Function: view
File: /var/www/html/public_html/application/controllers/Study.php
Line: 6544
Function: view
File: /var/www/html/public_html/index.php
Line: 315
Function: require_once
