PhysicsQ1–6 of 6 questions
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1
PhysicsMediumMCQIIT JEE · 1987
$A$ particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that:
A
It moves in a circular path
B
Acceleration is constant
C
Kinetic energy is constant
D
Both $(a)$ and $(c)$
Solution
(D) When a force acts perpendicular to the velocity of a particle,it does not change the magnitude of the velocity (speed),only its direction. This results in uniform circular motion.
Since the speed $v$ remains constant,the kinetic energy $K = \frac{1}{2}mv^2$ remains constant.
Because the force is always perpendicular to the velocity and the magnitude of the force is constant,the particle moves in a circular path.
Therefore,both statements $(a)$ and $(c)$ are correct.
Since the speed $v$ remains constant,the kinetic energy $K = \frac{1}{2}mv^2$ remains constant.
Because the force is always perpendicular to the velocity and the magnitude of the force is constant,the particle moves in a circular path.
Therefore,both statements $(a)$ and $(c)$ are correct.
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2
PhysicsMediumMCQIIT JEE · 1987
The displacement of a particle in a string stretched in the $X$ direction is represented by $y$. Among the following expressions for $y$,which ones describe wave motion?
A
$\cos kx \sin \omega t$
B
$k^2 x^2 - \omega^2 t^2$
C
$\cos (kx + \omega t)$
D
Both $(a)$ and $(c)$
Solution
(D) function $y(x, t)$ represents a wave motion if it satisfies the general wave equation $\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2}$.
$1$. For $y = \cos(kx + \omega t)$,this is a standard traveling wave equation of the form $f(kx \pm \omega t)$,which satisfies the wave equation.
$2$. For $y = \cos kx \sin \omega t$,this can be written as $\frac{1}{2} [\sin(kx + \omega t) - \sin(kx - \omega t)]$. This represents a standing wave,which is formed by the superposition of two traveling waves moving in opposite directions. Standing waves are also a form of wave motion.
Therefore,both expressions describe wave motion.
$1$. For $y = \cos(kx + \omega t)$,this is a standard traveling wave equation of the form $f(kx \pm \omega t)$,which satisfies the wave equation.
$2$. For $y = \cos kx \sin \omega t$,this can be written as $\frac{1}{2} [\sin(kx + \omega t) - \sin(kx - \omega t)]$. This represents a standing wave,which is formed by the superposition of two traveling waves moving in opposite directions. Standing waves are also a form of wave motion.
Therefore,both expressions describe wave motion.
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3
PhysicsMediumMCQIIT JEE · 1987
$A$ charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to:
A
$ - \frac{Q}{2} $
B
$ - \frac{Q}{4} $
C
$ + \frac{Q}{4} $
D
$ + \frac{Q}{2} $
Solution
(B) For the system of three charges to be in equilibrium,the net force on each charge must be zero.
Let the two charges $Q$ be placed at points $A$ and $B$ separated by a distance $x$. The charge $q$ is placed at the midpoint $C$ (distance $x/2$ from both $A$ and $B$).
Consider the equilibrium of charge $Q$ at point $B$. The force exerted by charge $Q$ at $A$ on charge $Q$ at $B$ must be balanced by the force exerted by charge $q$ at $C$ on charge $Q$ at $B$.
$F_{AB} + F_{CB} = 0$
$\frac{1}{4\pi\varepsilon_0} \frac{Q^2}{x^2} + \frac{1}{4\pi\varepsilon_0} \frac{qQ}{(x/2)^2} = 0$
$\frac{Q^2}{x^2} + \frac{4qQ}{x^2} = 0$
$Q^2 + 4qQ = 0$
$4qQ = -Q^2$
$q = -\frac{Q}{4}$
Let the two charges $Q$ be placed at points $A$ and $B$ separated by a distance $x$. The charge $q$ is placed at the midpoint $C$ (distance $x/2$ from both $A$ and $B$).
Consider the equilibrium of charge $Q$ at point $B$. The force exerted by charge $Q$ at $A$ on charge $Q$ at $B$ must be balanced by the force exerted by charge $q$ at $C$ on charge $Q$ at $B$.
$F_{AB} + F_{CB} = 0$
$\frac{1}{4\pi\varepsilon_0} \frac{Q^2}{x^2} + \frac{1}{4\pi\varepsilon_0} \frac{qQ}{(x/2)^2} = 0$
$\frac{Q^2}{x^2} + \frac{4qQ}{x^2} = 0$
$Q^2 + 4qQ = 0$
$4qQ = -Q^2$
$q = -\frac{Q}{4}$
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4
PhysicsEasyMCQIIT JEE · 1987
$A$ parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved further apart by means of insulating handles,then
A
The charge on the capacitor increases
B
The voltage across the plates decreases
C
The capacitance increases
D
The electrostatic energy stored in the capacitor increases
Solution
(D) When the battery is disconnected,the charge $q$ on the capacitor remains constant because there is no path for the charge to flow.
The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
When the distance $d$ between the plates is increased,the capacitance $C$ decreases.
Since $q = CV$,the voltage $V = \frac{q}{C}$ must increase as $C$ decreases.
The electrostatic energy stored in the capacitor is given by $U = \frac{q^2}{2C}$.
Since $q$ is constant and $C$ decreases,the energy $U$ increases.
The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
When the distance $d$ between the plates is increased,the capacitance $C$ decreases.
Since $q = CV$,the voltage $V = \frac{q}{C}$ must increase as $C$ decreases.
The electrostatic energy stored in the capacitor is given by $U = \frac{q^2}{2C}$.
Since $q$ is constant and $C$ decreases,the energy $U$ increases.
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5
PhysicsDifficultMCQIIT JEE · 1987
During the nuclear fusion reaction,
A
$A$ heavy nucleus breaks into two fragments by itself.
B
$A$ light nucleus bombarded by thermal neutrons breaks up.
C
$A$ heavy nucleus bombarded by thermal neutrons breaks up.
D
Two light nuclei combine to give a heavier nucleus and possibly other products.
Solution
(D) Nuclear fusion is a process in which two light atomic nuclei combine to form a single,heavier nucleus. This process is accompanied by the release of a large amount of energy due to the mass defect between the reactants and the product.
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6
PhysicsEasyMCQIIT JEE · 1987
During a negative beta decay,
A
An atomic electron is ejected.
B
An electron which is already present within the nucleus is ejected.
C
$A$ neutron in the nucleus decays emitting an electron.
D
$A$ part of the binding energy is converted into an electron.
Solution
(C) In negative $\beta$-decay,a neutron inside the nucleus transforms into a proton,an electron,and an antineutrino.
This process is represented by the equation:
$n \rightarrow p + e^- + \bar{\nu}_e$
Since the electron is created during the decay process of the neutron,option $C$ is correct.
This process is represented by the equation:
$n \rightarrow p + e^- + \bar{\nu}_e$
Since the electron is created during the decay process of the neutron,option $C$ is correct.
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