1
JEE Main 2015 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
JEE Main 2015 (Offline) Physics - Current Electricity Question 333 English
In the circuit shown, the current in the $$1\Omega $$ resistor is :
A
$$0.13$$ $$A,$$ from $$Q$$ to $$P$$
B
$$0.13$$ $$A$$, from $$P$$ to $$Q$$
C
$$1.3A$$ from $$P$$ to $$Q$$
D
$$0A$$
2
JEE Main 2015 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
A long cylindrical shell carries positives surfaces change $$\sigma $$ in the upper half and negative surface charge - $$\sigma $$ in the lower half. The electric field lines around the cylinder will look like figure given in :
(figures are schematic and not drawn to scale)
A
JEE Main 2015 (Offline) Physics - Electrostatics Question 245 English Option 1
B
JEE Main 2015 (Offline) Physics - Electrostatics Question 245 English Option 2
C
JEE Main 2015 (Offline) Physics - Electrostatics Question 245 English Option 3
D
JEE Main 2015 (Offline) Physics - Electrostatics Question 245 English Option 4
3
JEE Main 2015 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
In the given circuit, charges $${Q_2}$$ on the $$2\mu F$$ capacitor changes as $$C$$ is varied from $$1\,\mu F$$ to $$3\mu F.$$ $${Q_2}$$ as a function of $$'C'$$ is given properly by:
$$\left( {figures\,\,are\,\,drawn\,\,schematically\,\,and\,\,are\,\,not\,\,to\,\,scale} \right)$$

JEE Main 2015 (Offline) Physics - Capacitor Question 159 English
A
JEE Main 2015 (Offline) Physics - Capacitor Question 159 English Option 1
B
JEE Main 2015 (Offline) Physics - Capacitor Question 159 English Option 2
C
JEE Main 2015 (Offline) Physics - Capacitor Question 159 English Option 3
D
JEE Main 2015 (Offline) Physics - Capacitor Question 159 English Option 4
4
JEE Main 2015 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
A pendulum made of a uniform wire of cross sectional area $$A$$ has time period $$T.$$ When an additional mass $$M$$ is added to its bob, the time period changes to $${T_{M.}}$$ If the Young's modulus of the material of the wire is $$Y$$ then $${1 \over Y}$$ is equal to :
($$g=$$ $$gravitational$$ $$acceleration$$)
A
$$\left[ {1 - {{\left( {{{{T_M}} \over T}} \right)}^2}} \right]{A \over {Mg}}$$
B
$$\left[ {1 - {{\left( {{T \over {{T_M}}}} \right)}^2}} \right]{A \over {Mg}}$$
C
$$\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{A \over {Mg}}$$
D
$$\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{{Mg} \over A}$$

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