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Engineering
550 Views   |

100 g of water is heated from 30°C to 50°C. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J kg -1K-1)

• Option 1)

$4.2\: kJ$

• Option 2)

$8.4\: kJ$

• Option 3)

$84\: kJ$

• Option 4)

$2.1\: kJ$

As we learnt in Internal Energy - Fo an ideal gas E is a function of temperature only - wherein With change in temperature only kinetic energy is changed but not any potential energy.     Option 1) Incorrect  Option 2) Correct Option 3) Incorrect Option 4) Incorrect
Engineering
583 Views   |

A screw gauge gives the following reading when used to measure the diameter of a wire.

Main scale reading        :  0 mm

Circular scale reading   :  52 divisions

Given that 1 mm on main scale corresponds to 100 divisions of the circular scale.

The diameter of wire from the above data is :

• Option 1)

0.52 cm

• Option 2)

0.052 cm

• Option 3)

0.026 cm

• Option 4)

0.005 cm

As we learnt in  To measure the diameter of small spherical cylindrical body using Vernier Callipers - Vernier Constant = 1 Main scale division - 1 V.S. Division  V.C= 1 M.S.D - 1 V.S.D M.S.D=  Main Scale Reading V.S.D= Vernier Scale Reading   - wherein Total observed reading =  N= Nth division  Observations:  1.    Vernier constant (least count) of the Vernier Callipers:                     ...
Engineering
126 Views   |

A current $\dpi{100} I$ flows in an infinitely long wire with cross-section in the form of a semicircular ring of radius $\dpi{100} R$ The magnitude of the magnetic induction along its axis is

• Option 1)

$\frac{\mu _{0}I}{\pi ^{2}R}$

• Option 2)

$\frac{\mu _{0}I}{2\pi ^{2}R}$

• Option 3)

$\frac{\mu _{0}I}{2\pi R}$

• Option 4)

$\frac{\mu _{0}I}{4\pi R}$

As we learnt in  For Finite Length - - wherein    Current in a small element dI Magnetic field due to the element Option 1) Correct Option 2) Incorrect Option 3) Incorrect Option 4) Incorrect
Engineering
119 Views   |

The electrostatic potential inside a charged spherical ball is given by $\dpi{100} \phi = ar^{2}+b$ where $\dpi{100} r$ is the distance from the centre;     $\dpi{100} a,b$ are constants. Then the charge density inside the ball is

• Option 1)

$-24\pi a\varepsilon _{0}r$

• Option 2)

$-6 a\varepsilon _{0}r$

• Option 3)

$-24\pi a\varepsilon _{0}$

• Option 4)

$-6 a\varepsilon _{0}$

As we learnt in

Gauss's Law -

Total flux linked with a closed surface called Gaussian surface.

Formula:

$\dpi{100} \phi = \oint \vec{E}\cdot d\vec{s}=\frac{Q_{enc}}{\epsilon _{0}}$

- wherein

No need to be a real physical surface.

Qenc - charge enclosed by closed surface.

$\phi=ar^{2}+b$

Electric field   $E=\frac{-d\phi}{dr}=-2ar$ - (i)

$\phi=\int \vec{E}.d\vec{S}=\frac{q}{\varepsilon _{0}}$

$-2ar\:4\pi r^{2}=\frac{q_{in}}{\epsilon _{0}}$

$q_{inside}=-8\varepsilon _{0}a\pi\:r^{3}$

change density inside the ball is

$\rho _{inside}=\frac{q_{inside}}{\tfrac{4}{3}\pi\:r^{3}}$

$=\frac{-8\varepsilon _{0}a\pi\:r^{3}}{\tfrac{4}{3}\pi\:r^{3}}$

$\rho _{inside}=-6a\varepsilon _{0}$

Option 1)

$-24\pi a\varepsilon _{0}r$

This is incorrect option

Option 2)

$-6 a\varepsilon _{0}r$

This is incorrect option

Option 3)

$-24\pi a\varepsilon _{0}$

This is incorrect option

Option 4)

$-6 a\varepsilon _{0}$

This is correct option

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Engineering
133 Views   |

Two identical charged spheres suspended from a common point by two massless strings of length $\dpi{100} l$ are initially a distance $\dpi{100} d(d< < l)$ apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity $\dpi{100} \upsilon$ . Then as a function of distance $\dpi{100} x$ between them

• Option 1)

$\upsilon\: \alpha\: x^{-1/2}$

• Option 2)

$\upsilon\: \alpha\: x^{-1}$

• Option 3)

$\upsilon\: \alpha\: x^{1/2}$

• Option 4)

$\upsilon\: \alpha\: x$

Option 1 ans

Engineering
103 Views   |

A resistor $\dpi{100} R\: and\:\: 2\; \; \mu F$ capacitor in series is connected through a switch to $200 \: V$ direct supply. Across the capacitor is a neon bulb that lights up at $120 \: V$ Calculate the value of  $R$ to make the bulb light up $\dpi{100} 5 \: s$ after the switch has been closed.

$\dpi{100} \left ( \log_{10}2.5= 0.4 \right )$

• Option 1)

$1.3 \times 10^{4}\Omega$

• Option 2)

$1.7 \times 10^{5}\Omega$

• Option 3)

$2.7 \times 10^{6}\Omega$

• Option 4)

$3.3 \times 10^{7}\Omega$

As we learnt in  Current through the circuit Self Inductance - An emf is induced in the coil or the circuit which oppose the change that causes it. Which is also known back  emf. - wherein     Potential difference across capacitor Option 1) This option is incorrect Option 2) This option is incorrect Option 3) This option is correct Option 4) This option is incorrect
Engineering
105 Views   |

A fully charged capacitor  $\dpi{100} C$ with initial charge $\dpi{100} q_{0}$ is connected to a coil of self inductance $\dpi{100} L$ at $\dpi{100} t=0$ The time at which the energy is stored equally between the electric and the magnetic fields is

• Option 1)

$\pi \sqrt{LC}$

• Option 2)

$\frac{\pi}{4} \sqrt{LC}$

• Option 3)

$2\pi \sqrt{LC}$

• Option 4)

$\sqrt{LC}$

As we learnt in  LC Circuit voltage - - wherein    at t=0, maximum energy= When energy is equally distributed among capacitor and inductor then       or         Option 1) This option is incorrect Option 2) This option is correct Option 3) This option is incorrect Option 4) This option is incorrect
Engineering
175 Views   |

A boat is moving due east in a region where the earth’s magnetic field is 5.0 × 10-5 N A-1 m-1 due north and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 m s-1 , the magnitude of the induced emf in the wire of aerial is

• Option 1)

1 mV

• Option 2)

0.75 mV

• Option 3)

0.50 mV

• Option 4)

0.15 mV

As we learnt in  Velocity is perpendicular to magnetic field hence = Option 1) 1 mV This option is incorrect Option 2) 0.75 mV This option is incorrect Option 3) 0.50 mV This option is incorrect Option 4) 0.15 mV This option is correct
Engineering
110 Views   |

Two bodies of masses $\dpi{100} m\: and \: 4m$ are placed at a distance $\dpi{100} r$. The gravitational potential at a point on the line joining them where the gravitational field is zero is

• Option 1)

$zero$

• Option 2)

$-\frac{4Gm}{r}$

• Option 3)

$-\frac{6Gm}{r}$

• Option 4)

$-\frac{9Gm}{r}$

As we discussed in Gravitational field due to Point mass -     - wherein As the distance (r) of test mass from point (M) Increases I decreases.       Option 1) Incorrect option Option 2) Incorrect option Option 3) Incorrect option Option 4)
Engineering
198 Views   |

An object moving with a speed of $\dpi{100} 6.25 \: m\: s^{-1},$ is decelerated at a rate given by $\dpi{100} \frac{d\upsilon }{dt}= -2.5\sqrt{\upsilon }$

where $\dpi{100} \upsilon$ is the instantaneous speed. The time taken by the object, to come to rest, would be

• Option 1)

$1\: s$

• Option 2)

$2\: s$

• Option 3)

$4\: s$

• Option 4)

$8\: s$

As we learnt in Uniform velocity,Non uniform Velocity - Uniform velocity If equal displacement are covered in equal intervals of time. Non uniform velocity If equal displacement are covered in unequal intervals of time. - wherein More to know : For uniform motion along a straight line, the average and instantaneous velocities have the same values.       Option 1) Incorrect Option...
Engineering
119 Views   |

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is $\dpi{100} v$ , the total area around the fountain that gets wet is

• Option 1)

$\pi \frac{v^{2}}{g}$

• Option 2)

$\pi \frac{v^{4}}{g^{2}}$

• Option 3)

$\frac{\pi}{2} \frac{v^{4}}{g^{2}}$

• Option 4)

$\pi \frac{v^{2}}{g^{2}}$

As we learnt in  Horizontal Range - Horizontal distance travelled by projectile from the point of projectile to the point on ground where its hits. - wherein Special case of horizontal range For man horizontal range.     Option 1) Incorrect Option 2) Correct Option 3) Incorrect Option 4) Incorrect
Engineering
829 Views   |

A car is fitted with a convex side-view mirror of focal length 20 cm. A second car 2.8 m behind the first car is overtaking the first car at a relative speed of 15 m s-1. The speed of the image of the second car as seen in the mirror of the first one is

• Option 1)

$\frac{1}{10}ms^{-1}$

• Option 2)

$\frac{1}{15}ms^{-1}$

• Option 3)

$10\: ms^{-1}$

• Option 4)

$15\: ms^{-1}$

As we learnt in Mirror Formula -   - wherein Object distance from pole of mirror. Image distance from pole of mirror. focal length of the mirror.    From mirror formula so,   Correct option is 2.   Option 1) This is an incorrect option. Option 2) This is the correct option. Option 3) This is an incorrect option. Option 4) This is an incorrect option.
Engineering
262 Views   |

Let the $\dpi{100} x-z$ plane be the boundary between two transparent media. Medium 1 in $\dpi{100} z\geq 0$ has a refractive index of $\dpi{100} \sqrt{2}$ and medium 2 with $\dpi{100} z< 0$ has a refractive index of $\dpi{100} \sqrt{3}$. . A ray of light in medium 1 given by the vector $\dpi{100} \vec{A}= 6\sqrt{3}\: \hat{i}+8\sqrt{3}\: \hat{j}-10\: \hat{k}$ is incident on the plane of separation. The angle of refraction in medium 2 is

• Option 1)

$30^{\circ}$

• Option 2)

$45^{\circ}$

• Option 3)

$60^{\circ}$

• Option 4)

$75^{\circ}$

As we learnt in Relation between angle of incidence and angle of refaction -   - wherein refractive index of medium of incidence.   refractive index of medium where rays is refracted. angle of incidence. angle of refraction.     From Snell's law Correct option is 2.   Option 1) This is an incorrect option. Option 2) This is the correct option. Option 3) This is an incorrect...
Engineering
121 Views   |

Direction : The question has a paragraph followed by two statements, Statement-1 and Statement-2. Of the given four alternatives after the statements, choose the one that describes the statements.

A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film.

Statement - 1: When light reflects from the air- glass plate interface, the reflected wave suffers a phase change of $\dpi{100} \pi$.

Statement - 2: The centre of the interference pattern is dark.

• Option 1)

Statement-1 is true, Statement-2 is false.

• Option 2)

Statement-1 is true , Statement-2 is true, Statement-2 is the correct explanation of Statement-1.

• Option 3)

Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of Statement-1.

• Option 4)

Statement-1 is false, Statement-2 is true.

As we learnt in Relation between angle of incidence and angle of refaction -   - wherein refractive index of medium of incidence.   refractive index of medium where rays is refracted. angle of incidence. angle of refraction.     Thin lens formula -   - wherein are object and image distance from lens.   S1= When light reflects from denser (glass) a phase drift of  is generated. S2=...
Engineering
97 Views   |

A mass $\dpi{100} M$ attached to a horizontal spring, executes $\dpi{100} SHM$ with a amplitude $\dpi{100} A_{1}$ When the mass$\dpi{100} M$ passes through its mean position then a smaller mass $\dpi{100} m$ is placed over it and both of them move together with amplitude $\dpi{100} A_{2}$. The ratio of $\dpi{100} \left ( \frac{A_{1}}{A_{2}} \right )$ is

• Option 1)

$\frac{M}{M+m }$

• Option 2)

$\frac{M+m}{M }$

• Option 3)

$\left ( \frac{M}{M+m } \right )^{1/2}$

• Option 4)

$\left ( \frac{M+m}{M } \right )^{1/2}$

As we learnt in Time period of oscillation for spring mass system - - wherein m = mass of block K = spring constant    From conservation of linear momentum                                           Correct option is 4.   Option 1) This is an incorrect option.   Option 2) This is an incorrect option. Option 3) This is an incorrect option. Option 4) This is the correct option.
Engineering
129 Views   |

The transverse displacement $\dpi{100} y\left ( x,t \right )$ of a wave on a string is given by

$\dpi{100} y\left ( x,t \right )= e^{-\left ( ax^{2}+bt^{2} +2\sqrt{abxt}\right )}$ This represents a

• Option 1)

$wave\: moving\: in\: +x-direction\: with \: speed \sqrt{\frac{a}{b}}$

• Option 2)

$wave\: moving\: in\: -x\: -direction\: with \: speed \sqrt{\frac{b}{a}}$

• Option 3)

$standing\: wave\: of\: f\! requency\:\sqrt{b}$

• Option 4)

$standing\: wave\: of\: f\! requency\:\frac{1}{\sqrt{b}}$

As we learnt in Relation between phase difference and path difference - Phase difference -     It is function of type   Wave velcoity Correct option is 2.   Option 1) This is an incorrect option. Option 2) This is the correct option. Option 3) This is an incorrect option. Option 4) This is an incorrect option.
Engineering
104 Views   |

Two particles are executing simple harmonic motion of the same amplitude A and frequency $\dpi{100} \omega$ along the $\dpi{100} x$-axis. Their mean position is separated by distance $\dpi{100} X_{0}(X_{0}> A)$. If the maximum separation between them is $\dpi{100} (X_{0}+A)$, the phase difference between their motion is

• Option 1)

$\frac{\pi }{2}$

• Option 2)

$\frac{\pi }{3}$

• Option 3)

$\frac{\pi }{4}$

• Option 4)

$\frac{\pi }{6}$

As we learnt in Simple harmonic as projection of circular motion - - wherein     The crossing happens when  Correct option is 2.   Option 1) This is an incorrect option. Option 2) This is the corect option. Option 3) This is an incorrect option. Option 4) This is an incorrect option.
Engineering
705 Views   |

Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly ( Surface tension of soap solution = 0.03 N m-1)

• Option 1)

$4\pi \: mJ$

• Option 2)

$0.2\pi \: mJ$

• Option 3)

$2\pi \: mJ$

• Option 4)

$0.4\pi \: mJ$

As we learnt in Surface Energy - - wherein     Correct option is 4. Option 1) This is an incorrect option. Option 2) This is an incorrect option. Option 3) This is an incorrect option. Option 4) This is the correct option.
Engineering
864 Views   |

Water is flowing continuously from a tap having an internal diameter 8 × 10-3 m . The water velocity as it leaves the tap is 0.4 m s-1 . The diameter of the water stream at a distance 2 × 10-1 m below the tap is close to

• Option 1)

5.0 × 10-3 m

• Option 2)

7.5 × 10-3 m

• Option 3)

9.6 × 10-3 m

• Option 4)

3.6 × 10-3 m

As we learnt in Equation of Continuity - Mass of the liquid entering per second at A = mass of the liquid leaving per second at B.  a1 v1 = a2 v2  - wherein a1  and a2 be the area of cross section.         &       Corrrect option is 4. Option 1) 5.0 × 10-3 m This is an incorrect option. Option 2) 7.5 × 10-3 m This is an incorrect option. Option 3) 9.6 × 10-3 m This is an incorrect...
Engineering
127 Views   |

The half life of a radioactive substance is 20 minutes. The approximate time interval $\dpi{100} \left ( t_{2}-t_{1} \right )$ between the time $\dpi{100} t_{2}$ when $\dpi{100} \frac{2}{3}$ of it has decayed and time $\dpi{100} t_{1}$ when $\dpi{100} \frac{1}{3}$ of it had decayed is

• Option 1)

$7 \: min$

• Option 2)

$14 \: min$

• Option 3)

$20 \: min$

• Option 4)

$28 \: min$

As we learnt in Number of nuclei after disintegration -  or  - wherein Number of nucleor activity at a time is exponentional function     at t = t2,   or                                              (1) at  t = t1,   or                                              (2) Divide (1) in (2)     or    or    or  Correct option is 3. Option 1) This is an incorrect option. Option 2) This is an...
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