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13 views4 pages

RT - 1-1

Rt1

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santhiyaarul27
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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‭ II‬‭th‬ ‭std‬

X ‭REVESION TEST – 1‬
‭Time : 1Hr 45 minutes PHYSICS [ NCERT + NCERT EXEMPLER ] M.M : 50‬
‭TEST TOPIC :‬‭Electric Charges and Fields , Electrostatic Potential and Capacitance , Current Electricity .‬

‭PART-A‬ ‭.‬

‭1.‬ ‭(A)‬‭Show, on a plot, variation of resistivity as a function of temperature.‬


‭(i)‬‭a conductor‬ ‭(a)‬ ‭copper‬ ‭(b)‬‭nichrome‬‭(ii)‬‭a typical semiconductor (gallium arsenide)‬
‭(B)‬ ‭Using the expression for the resistivity in terms‬‭of number density and relaxation time between the collisions,‬
‭explain how resistivity in the case of a conductor increases while it decreases in a semiconductor, with the rise of‬
‭Temperature.‬
‭(C)‬ ‭Define the terms 'temperature coefficient of resistance' and 'electrical conductivity of a conductor.‬
‭Why are constantan and manganin used for making standard resistances? Explain.‬

‭2.‬ T
‭ wo conducting spheres of radius r‬‭1‬ ‭= 8 cm and r‬‭2‬ ‭= 2 cm are separated by a distance much larger than 8 cm and are‬
‭connected by a thin conducting wire as shown in the‬‭fig – 1‬‭. A total charge of Q = +100 nC is placed on one of the‬
‭spheres.After a fraction of a second, the charge Q is redistributed and both the spheres attain electrostatic equilibrium.‬

‭fig – 1‬

(‭ i)‬‭Calculate the charge and surface charge density on each sphere.‬


‭(ii)‬‭Calculate the potential at the surface of each sphere.‬

‭3.‬ (‭ A)‬‭Define DIELECTRICS AND POLARISATION .‬


‭(B)‬‭Explain the difference in behaviour of a conductor and a dielectric in an external electric field.‬
‭(C)‬‭Show that‬‭a dielectric develops a net dipole moment in an external electric field.‬
‭(a) Non-polar molecules, (b) Polar molecules. Using diagram‬
‭(D)‬ ‭Differentiate between (a) Non-polar molecules, (b) Polar molecules & give 2 example of each‬
‭(E)‬‭Write the relation between polarization of a dielectric with electric susceptibility.‬

‭PART-B‬ ‭.‬

‭4.‬ A
‭ conducting spherical shell of radius R has charge Q uniformly distributed on them. The centre of the sphere is at‬
‭x = 0 . Find the magnitude and direction of the‬‭net electric field and potential‬ ‭on the X-axis at‬
‭𝑅‬
‭(i) x = 0 (ii) x =‬ ‭(iii) x = 3R . Plot relation between‬‭𝐸‬ ‭Vs r. For given case‬
‭2‬

‭5.‬ ‭Find currents in different branches of the electric circuit shown in‬‭fig – 2‬‭.‬

‭fig – 2‬
‭6.‬ ‭(A)‬‭Describe the behavior of a dipole when placed at different angles with respect to a uniform electric field‬
‭(i) θ = 0° (ii) θ = 90° (iii) θ = 180°‬
‭(B)‬‭An electric dipole of length 4 cm, when placed‬‭with its axis making an angle of 60° with a uniform‬
‭electric field, experiences a torque of‬ ‭4‬ ‭3‬‭Nm. Calculate the potential energy of the dipole,‬
‭if it has charge ± 8 nC .‬

‭7.‬ ‭Find the expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d when a‬
‭‬
𝑑 ‭‬
𝑑
‭dielectric slab of thickness‬‭𝑡‬ = ‭4‬
‭& a metallic slab of thickness‬ ‭𝑡‬ = ‭3‬
‭where (‬‭𝑡‬‭< d) are introduced one by‬
‭one between the plates of the capacitor. In which case would the capacitance be more and why?‬

‭PART-C‬ ‭.‬

‭8.‬ F
‭ ind the emf and internal resistance of a single battery which is equivalent to a combination of three batteries as‬
‭shown in‬‭fig – 3‬‭.‬

‭fig – 3‬

‭9.‬ O
‭ hm's law is not a fundamental law of nature. State the conditions under which Ohm's law fails. Give one example‬
‭for each case.‬

‭10.‬ ‭Two charges q1 and q2 are placed at (0, 0, d) and (0, 0, –d) respectively. Find the locus of points where the potential‬
‭is zero.‬

‭11.‬ ‭The electric field in a region is given by‬ ‭𝐸‬‭‬ = ‭‬(‭10‬‭‬‭𝑥‬‭‬ + ‭‬‭4‬)‭‬‭𝑖‬
‭where‬‭𝑥‬‭is in meter and E is in N/C. Calculate the amount of work done in taking a unit charge from‬
‭(i) (5 m, 0) to (10 m, 0) (ii) (5 m, 0) to (5 m, 10 m)‬

‭12.‬ ‭Two charges 2‬µ‭C and –2‬µ‭C are placed at points A and B 6 cm apart.‬
‭(A‬‭) Identify an equipotential surface of the system.‬
‭(B)‬‭What is the direction of the electric field at every point on this surface?‬

‭13.‬ ‭The‬‭fig – 4‬ ‭shows electric potential V as a function of‬‭x‬‭.‬

‭fig – 4‬

‭(i)‬ ‭Rank the four regions according to the magnitude of‬ ‭x‬‭- component of the electric field E‬
‭(ii)‬‭Plot electric field (E) as a function of‬ ‭x ,‬‭for the above figure .‬

‭PART-D‬ ‭.‬

‭Assertion-Reason Type Questions‬‭(14 - 16)‬


‭(A)‬‭Both Assertion (A) and Reason (R) are true but Reason (R) is‬‭NOT‬‭correct explanation of Assertion (A)‬
(‭ B)‬‭Assertion (A) is true but Reason (R) is false‬
‭(C)‬‭Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)‬
‭(D)‬‭Assertion (A) is false and Reason (R) is true‬‭.‬

‭14.‬ ‭Assertion (A) :‬ ‭As per the‬‭Fig – 5‬‭, a point charge +q is placed at origin O. Work done in taking another point‬
‭charge – Q from the point A (0, a) to another point B (a, 0) along the straight path AB is zero‬

‭Fig – 5‬

‭Reason (R) :‬‭Work done to bring charge – Q from A to point B is given by , W =‬− ( ‭𝑘‬(−‭𝑄)‬
‭𝑎‬

‭𝑘‬(−‭𝑄)‬
‭𝑎‬ )
‭15.‬ ‭Assertion (A) :‬ ‭If the amount of charge flowed in time t through a cross-section of wire is‬‭q = βt − γt‬‭2‬ ‭, where β and‬
‭γ are constants.‬
‭‬
1 ‭2‬ ‭‬
1 ‭3‬
‭Reason (R) :‬ ‭The current `i` in terms of t is given by‬‭𝑖‬ = ‭2‬
‭β‭𝑡‬ ‬ − ‭3‬ ‭γ‭𝑡‬ ‬ ‭.‬

‭16.‬ ‭Assertion (A) :‬ ‭Electric field is a characteristic of the system of charges and is independent of the test charge that‬
‭you place at a point to determine the field.‬
‭Reason (R) :‬‭The magnitude of electric field E due to a point charge is thus same on a sphere with the point charge‬
‭at its centre; in other words, it has a spherical symmetry.‬

‭17.‬ ‭Refer to the arrangement of charges in Fig – 6 . and a Gaussian surface of radius R with Q at the centre. Then‬
‭Select the correct entity‬

‭Fig – 6‬

−‭𝑄‬ −‭𝑄‬
‭(I)‬‭total flux through the surface of the sphere is‬ ‭(II)‬‭field on the‬‭surface of the sphere is‬ ‭2‬
ε‭0‬ ‭4π
‬ ε‭0‭𝑅 ‬

‭(III)‬‭flux through the surface of sphere due to 5Q is zero.‬ ‭(IV)‬‭field on the‬‭surface of sphere due to –2Q is same‬
‭Everywhere.‬

‭(A) II & IV Only (B) I & IV Only (C) I & III Only (D) I & II Only‬

‭18.‬ ‭A hemisphere is uniformly charged positively. The electric field at a point on a diameter away from the centre‬
‭is directed‬
‭(A) perpendicular to the diameter (B) parallel to the diameter‬
‭(C) at an angle tilted towards the diameter (D) at an angle tilted away from the diameter.‬

‭19.‬ ‭According to this diagram, the potential difference across the terminals is (internal resistance of cell = r)‬

‭(A) V = E – i r (B) V = E + i r (C) V = E (D) Zero‬


‭1‬
‭20.‬ ‭A metal rod of length 10 cm and a rectangular cross-section of 1cm ×‬ ‭2‬ ‭cm is connected to a battery across‬
‭ pposite faces. The resistance will be‬
o
‭1‬
‭(A) maximum when the battery is connected across 1 cm ×‬ ‭2‬ ‭cm faces.‬
(‭ B) maximum when the battery is connected across 10 cm × 1 cm faces.‬
‭1‬
‭(C) maximum when the battery is connected across 10 cm ×‬ ‭2‬ ‭cm faces.‬
‭(D) same irrespective of the three faces.‬

‭21.‬ ‭Two batteries of emf ε‬‭1‬ ‭and ε‬‭2‬ ‭(ε‬‭2‬ ‭> ε‬‭1‬‭) and internal resistances r‬‭1‬ ‭and r‬‭2‬ ‭respectively are connected in parallel‬
‭as shown in Fig 7‬

‭Fig 7‬

(‭ A) The equivalent emf ε‬‭eq‬ ‭of the two cells is between‬‭ε‭1‬ ‬ ‭and ε‬‭2‬‭, i.e. ε‬‭1‬‭< ε‬‭eq‬ ‭< ε‬‭2‭.‬ ‬
‭(B) The equivalent emf εeq is smaller than ε‬‭1‬ ‭.‬
‭(C) The ε‬‭eq‬ ‭is always given by ε‬‭eq‬ ‭= ε‬‭1‬ ‭+ ε‬‭2‬ ‭always.‬
‭(D) εeq is independent of internal resistances r‬‭1‬ ‭and r‬‭2‭.‬ ‬

‭22.‬ ‭A charge q is to be distributed on two conducting spheres. What should be the value of the charges on the‬
‭spheres so that the repulsive force between them is maximum when they are placed at a fixed distance from‬
‭each other in air?‬
‭𝑞‬ ‭𝑞‬ ‭𝑞‬ ‭3‭𝑞
‬‬ ‭𝑞‬ ‭2‭𝑞
‬‬ ‭𝑞‬ ‭4‭𝑞
‬‬
‭(A)‬ ‭and‬ ‭(B)‬ ‭and‬ ‭(C)‬ ‭and‬ ‭(D)‬ ‭and‬
‭2‬ ‭2‬ ‭4‬ ‭4‬ ‭3‬ ‭3‬ ‭5‬ ‭5‬

‭23.‬ ‭Fig – 8 shows the variation of electric field intensity E versus distance x. What is the potential difference‬
‭between the points at x = 2 m and at x = 6 m from O?‬

‭Fig – 8‬

‭(A) 30 V (B) 60 V (C) 40 V (D) 80 V‬

‭24.‬ ‭The effective capacitance of combination of combination of equal capacitors between points A and B shown in‬
‭Fig – 9 is‬

‭Fig – 9‬

‭‬
𝐶 ‭‬
𝐶
‭(A)‬ ‭(B)‬ ‭(C)‬ ‭2‬‭𝐶‬ ‭(D)‬ ‭4‬‭𝐶‬
‭2‬ ‭4‬

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