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When a galvanometer is shunted with a 4 resistance, the deflection is reduced to onefifth. If the galvanometer is further shunted with a 2 wire, the further reduction in the deflection will be (the main current remains the same).
An inductor (x_{L} = 2) a capacitor (x_{C} = 8) and a resistance (8) is connected in series with an A.C. source. The voltage output of A.C. source is given by v = 10 cos 100πt. The instantaneous potential difference between A and B, when it is half of the voltage output from source at that instant will be :
Figure shows a system of three concentric metal shells A, B and C with radii a, 2a and 3a respectively. Shell B is earthed and shell C is given a charge Q. Now if shell C is connected to shell A, then the final charge on the shell B, is equal to :
A ring of mass m, radius r having charge q uniformly distributed over it and free to rotate about its own axis is placed in a region having a magnetic field B parallel to its axis. If the magnetic field is suddenly switched off, the angular velocity acquired by the ring is :
Four infinite ladder network containing identical resistances of R each, are combined as shown in figure. The equivalent resistance between A and B is R_{AB} and between A and C is R_{AC}. Then the value of is :
Let the equivalent resistance of one infinite ladder be x. Then the complete network reduces to
Two small balls, each having equal positive charge Q are suspended by two insulating strings of equal length L from a hook fixed to a stand. If the whole setup is transferred to a satellite in orbit around the earth, the tension in equilibrium in each string is equal to
A satellite is in a state of free fall & hence weightlessness. Thus only electric force is responsible for the tension
In a practical wheat stone bridge circuit as shown, when one more resistance of 100 ? is connected in parallel with unknown resistance ' x ', then ratio l_{1}/l_{2} becomes '2'. l_{1} is balance length. AB is a uniform wire. Then value of ' x ' must be :
Loop A of radius (r << R) moves towards loop B with a constant velocity V in such a way that their planes are always parallel. What is the distance between the two loops (x) when the induced emf in loop A is maximum
The figure shows, a graph of the current in a discharging circuit of a capacitor through a resistor of resistance 10.
A single circular loop of wire with radius 0.02 m carries a current of 8.0 A. It is placed at the centre of a solenoid that has length 0.65 m, radius 0.080 m and 1300 turns.
In front of an earthed conductor a point charge + q is placed as shown in figure :
Charge is distributed over the surface of conductor in such a way that net field due to this charge and outside charge q is zero inside the conductor. Field due to only q is nonzero.
In the figure shown the key is switched on at t = 0. Let I_{1} and I_{2} be the currents through inductors having self inductances L_{1} & L_{2} at any time t respectively. The magnetic energy stored in the inductors 1 and 2 be U_{1} and U_{2}. Then at any instant of time is :
Statement – 1
Four point charges q_{1}, q_{2}, q_{3} and q_{4} are as shown in figure. The flux over the shown Gaussian surface depends only on charges q_{1} and q_{2}.
Statement – 2
Electric field at all points on Gaussian surface depends only on charges q_{1} and q_{2}.
Statement ? is true directly from Gauss Theorem.
Statement 2 is false. Electric field at any point an Gaussian surface depends on all four charges.
Statement1 is True, Statement2 is False
Statement 1 : A direct uniformly distributed current flows through a solid long metallic cylinder along its length. It produces magnetic field only outside the cylinder .
Statement 2 : A thin long cylindrical tube carrying uniformly distributed current along its length does not produce a magnetic field inside it. Moreover, a solid cylinder can be supposed to be made up of many thin cylindrical tubes.
The current through solid metallic cylinder also produces magnetic field inside the cylinder. Hence statement1
is false
Statement–1 : Magnitude of potential difference across the terminals of a nonideal battery in a circuit cannot be greater than its emf.
Statement–2 : When a current of magnitude I is passing through a battery of emf E and internal resistance r as shown, the magnitude of potential difference (V) across the battery is given by V = E– I r
Statement1 is obviously false if the current is sent in opposite direction given in figure of statement2
Statement–1 : No electric current will be present within a region having uniform and constant magnetic field.
Statement–2 : Within a region of uniform and constant magnetic field , the path integral of magnetic field along any closed path is zero. Hence from Ampere circuital law
(where the given terms have usual meaning), no current can be present within a region having uniform and constant magnetic field.
along any closed path within a uniform magnetic field is always zero. Hence the closed path can be chosen of any size, even very small size enclosing a very small area. Hence we can prove that net current through each area of infinitesimally small size within region of uniform magnetic field is zero. Hence we can say no current (rather than no net current) flows through region of uniform magnetic field. Hence statement 2 is correct explanation of statement1.
In the circuit given below, both batteries are ideal. EMF E_{1} of battery 1 has a fixed value, but emf E_{2} of battery 2 can be varied between 1.0 V and 10.0 V. The graph gives the currents through the two batteries as a function of E_{2}, but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite the direction of that battery's emf. (Direction of emf is from negative to positive)
Q. The value of emf E_{1} is :
In the circuit given below, both batteries are ideal. EMF E_{1} of battery 1 has a fixed value, but emf E_{2} of battery 2 can be varied between 1.0 V and 10.0 V. The graph gives the currents through the two batteries as a function of E_{2}, but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite the direction of that battery's emf. (Direction of emf is from negative to positive)
Q. The resistance R_{1} has value :
In the circuit given below, both batteries are ideal. EMF E_{1} of battery 1 has a fixed value, but emf E_{2} of battery 2 can be varied between 1.0 V and 10.0 V. The graph gives the currents through the two batteries as a function of E_{2}, but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite the direction of that battery's emf. (Direction of emf is from negative to positive)
Q. The resistance R_{2} is equal to :
Curves in the graph shown give, as functions of radial distance r (from the axis), the magnitude B of the magnetic field (due to individual wire) inside and outside four long wires a, b, c and d, carrying currents that are uniformly distributed across the cross sections of the wires. Overlapping portions of the plots are indicated by double labels. All curves start from the origin.
Q. Which wire has the greatest radius ?
Curves in the graph shown give, as functions of radial distance r (from the axis), the magnitude B of the magnetic field (due to individual wire) inside and outside four long wires a, b, c and d, carrying currents that are uniformly distributed across the cross sections of the wires. Overlapping portions of the plots are indicated by double labels. All curves start from the origin.
Q. Which wire has the greatest magnitude of the magnetic field on the surface ?
Curves in the graph shown give, as functions of radial distance r (from the axis), the magnitude B of the magnetic field (due to individual wire) inside and outside four long wires a, b, c and d, carrying currents that are uniformly distributed across the cross sections of the wires. Overlapping portions of the plots are indicated by double labels. All curves start from the origin.
Q. The current density in wire a is
Match the following :
The following table gives the lengths of four copper rods at the same temperature, their diameters, and the potential differences between their ends.
Correctly match the physical quantities mentioned in the left column with the rods as marked.
(A) Rod 1 (p) Greatest Drift speed of the electrons.
(B) Rod 2 (q) Greatest Current
(C) Rod 3 (r) Second greatest current
(D) Rod 4 (s) Greatest Electric field
(t) greatest resistance
A square loop of uniform conducting wire is as shown in figure. A currentI ( in amperes) enters the loop from one end and exits the loop from opposite end as shown in figure. The length of one side of square loop is l metre. The wire has uniform cross section area and uniform linear mass density. In four situations of columnI, the loop is subjected to four different uniform and constant magnetic field. Under the conditions of columnI, match the columnI with corresponding results of columnII ( B_{o} in column I is a positive nonzero constant)
In an f.c.c. crystal, which of the following shaded planes contains the following type of arrangement of atoms?
Shown arrangement is hexagonally closed pack plane and these type of planes are arranged perpendicular to
body diagonal of fcc unit cell as shown.
In an f.c.c. unit cell, atoms are numbered as shown below. The atoms not touching each other are :
(Atom numbered 3 is face centre of front face).
Atoms along one edge or at corners do not touch each other in fcc cell.
Electrolysis of a solution of HSO_{4}^{} ions produces S_{2}O_{8}^{– –}. Assuming 75% current efficiency, what current should be employed to achieve a production rate of 1 mole of S_{2}O_{8}^{ } per hour ?
You are given the following cell at 298 K, = – 0.76 V. Which of the following amounts of NaOH (equivalent weight = 40) will just make the pH of cathodic compartment to be equal to 7.0 :
For the following parallel chain reaction
what will be that value of overall halflife of A in minutes?
Decomposition of A follows first order kinetics by the following equation.
4A(g) ——> B(g) + 2C(g)
If initially, total pressure was 800 mm of Hg and after 10 minutes it is found to be 650 mm of Hg. What is halflife of A ? (Assume only A is present initially)
Assuming the formation of an ideal solution, determine the boiling point of a mixture containing 1560 g benzene (molar mass = 78) and 1125 g chlorobenzene (molar mass = 112.5) using the following against an external pressure of 1000 Torr:
A complex containing K^{+}, Pt(IV) and Cl^{—} is 100% ionised. Given that i = 3. Thus, complex is :
and in [PtCl_{6}]^{2}ion oxidation state of Pt is +4.
A current of 2.68 A is passed for one hour through an aqueous solution of CuSO_{4} using copper electrodes . Select the correct statement(s) from the following :
Increase in mass of cathode = decrease in mass of Anode =
Decomposition of 3 A(g) ——> 2 B(g) + 2C(g) follows first order kinetics. Initially only A is present in the container. Pressure developed after 20 min. and infinite time are 3.5 and 4 atm respectively. Which of the following is true.
Which of the following is/are correct for an ideal binary solution of two volatile liquids (eg. benzene and toluene)?
Statement1 : Time taken for the completion of 75% of a Ist order reaction is double than its t_{½}.
Statement2 : Time taken for completion of any fraction of a Ist order reaction is proportional to the extent of reaction completed.
Statement1 is correct but statement2 is incorrect. The time taken for completion of any fraction of a Ist order
reaction is a fixed maximum value for a particular reaction.
Statement1 : The difference in the boiling points of equimolar solution of HCl and HF decreases as their molarity is decreased.
Statement2 : The extent of dissociation decreases steadily with increasing dilution.
Statement1 is correct because As molarity decreases degree of dissociation of both HCI and HF will increase and approach unity.
Statement2 is false because degree of dissociation increases with dilution.
Statement1 : A gas with higher critical temperature is absorbed more than a gas with lower critical temperature on the same absorbent.
Statement2 : Higher critical temperature implies that the gas is more easily liquifiable.
Both the statements are individually correct but they are not related.
Statement1 : Specific conductance decreases with dilution whereas equivalent conductance increases.
Statement2 : On dilution number of ions per millilitre decreases but total number of ions increases considerably.
Total number of ions will increase slightly on dilution (Not considerably).
The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation
When a certain conductivity cell (C) was filled with 25 x 10^{–4} (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl^{–} and SO_{4}^{–2} are 80 ohm^{–1} cm^{2} mole^{–1 }and 160 ohm^{–1} cm^{2} mole^{–1} respectively.
Q. What is the molar conductance of NaCl at infinite dilution ?
The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation
When a certain conductivity cell (C) was filled with 25 x 10^{–4} (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl^{–} and SO_{4}^{–2} are 80 ohm^{–1} cm^{2} mole^{–1 }and 160 ohm^{–1} cm^{2} mole^{–1} respectively.
Q. What is the cell constant of the conductivity cell (C) ?
The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation
When a certain conductivity cell (C) was filled with 25 x 10^{–4} (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl^{–} and SO_{4}^{–2} are 80 ohm^{–1} cm^{2} mole^{–1 }and 160 ohm^{–1} cm^{2} mole^{–1} respectively.
Q. If the cell (C) is filled with 5 x 10^{–3} (N) Na_{2}SO_{4} the obserbed resistance was 400 ohm. What is the molar conductance of Na_{2}SO_{4} ?
Properties, whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionisation or association hence these properties are also affected. Number of moles of the product is related to degree of ionisation or association by vant Hoff factor ‘i’ given by i = [ 1 + (n – 1) α] for dissociation where n is the number of products (ions or molecules) obtained per mole of the reactant & i = for association.
Where n is number of reactant particles associated to give 1 mole product.
A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant K_{b}. The solute dimerises in the solution according to the following equation. The degree of association is α.
Q. The vant Hoff factor will be [ if we start with one mole of X ]
Properties, whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionisation or association hence these properties are also affected. Number of moles of the product is related to degree of ionisation or association by vant Hoff factor ‘i’ given by i = [ 1 + (n – 1) α] for dissociation where n is the number of products (ions or molecules) obtained per mole of the reactant & i = for association.
Where n is number of reactant particles associated to give 1 mole product.
A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant K_{b}. The solute dimerises in the solution according to the following equation. The degree of association is α.
Q. The colligative properties observed will be
(A) During association
Properties, whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionisation or association hence these properties are also affected. Number of moles of the product is related to degree of ionisation or association by vant Hoff factor ‘i’ given by i = [ 1 + (n – 1) α] for dissociation where n is the number of products (ions or molecules) obtained per mole of the reactant & i = for association.
Where n is number of reactant particles associated to give 1 mole product.
A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant K_{b}. The solute dimerises in the solution according to the following equation. The degree of association is α.
Q. The equilibrium constant for the process can be expressed as
For A + B → C in column  II the graphs given can be from any of these four types.
(p) – Vs time (x axis) (q) t_{1/2} Vs initial concentration (x axis) (s) Conc. Vs time (x axis) (t) log k Vs 1/Temperature (x axis)
Match the graphs in Column–II for the given order of reactions in Column – I
(A) First order :– t_{1/2} is constant. (p)
concentration and rate decrease exponentially.(q)
Arhennius equation is applicable for every order reaction (t)
(B) Zero order : Rate is constant. (p)
Concentration decreases linearly with t. (r)
Arhennius equation is applicable for every order reaction (t)
(C) Second order :
Arhennius equation is applicable for every order reaction (t)
(D) Same as (A).
Two roads OA and OB intersect at an angle of 60º. A car driver approaches O from A where OA = 800m at a uniform speed of 20m/s. Simultaneously another car moves from O towards B at a uniform speed of 25m/s. If t is time when two cars are closest, find t.
Let the distance between two cars after t seconds be s. Distance covered by first car in t second is 20t meter, so its distance from O will be (800 – 20t) meter and distance covered by second car will be 25t meter.
If f(x) satisfies x + f (x) = 2f (x) then f^{–1} (x) satisfies
Let f : R → R and g : R → R be two one  one and onto functions such that they are mirror images of each other about the line y = 0, then h (x) = f(x) + g(x) is
f : R → R & g : R →R be two oneone onto functions such that f & g are mirror images of each other about line y = 0. It means one is –ve of the other
i.e. f(x) = – g(x)
⇒ f(x) + g(x) = 0
⇒ h(x) = 0
h(x) is not onto as well as not oneone
The function f (x) = (x^{2} – 1) x^{2} – 3x + 2 + cos x is not differentiable at x =
The function f (x) = (x – [x]) sin p x is, (where [.] denotes greatest integer function)
f(x) is continuous at every integer & also f(x) is continuous for any other real number. So f(x) is continuous for all real numbers.To check differentiability
If f(x) = x^{3} + ax^{2} + ax + x(tanθ + cotθ) is increasing for all real x and if then
For what values of a, m and b, Lagrange's mean value theorem is applicable to the function f(x) for x [0, 2]
Let f(x) = Then f(x) is strictly increasing in R for
If f is continuous at a point. Then f will also continuous at that point.
If f is differentiable then ff will also be differentiable
Statement1 : If a 4^{th} degree polynomial function has four distinct real roots, then its graph have exactly two inflection points.
Statement2 : If the equation f"(x) = 0 has distinct real roots, then the equation f(x) = 0 has atleast four real roots.
If a 4th degree polynomial function has four distinct real roots, then its graph will be like following
From graph we can say curve has 2 point of inflection.
Statement 1 is true
Statement 2 is not true.
Statement1 : Let 0 < a < b. Then Rolle's theorem is applicable to the function f(x) = in [a,b] and Rolle's constant is the GM of a and b.
Statement2 : All the three conditions of Rolle's theorem are satisfied by the above function f(x).
Statement1 All points of intersection of y = f (x) & y = f ^{– 1} (x) lies on y = x only.
Statement2 If point P (α, β) lies on y = f (x) then Q (β, α) lies on y = f ^{– 1} (x).
Pt. of intersection of y = f(x) & y = f^{1}(x) lies on y = x. Also they may not lie on y = x. Infact points which does not
lies on y = x are even in numbers.
and statement (2) is true.
Statement1 where [.] represent greatest integer function.
Statement2
The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation f(x) + g(x) = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.
Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f^{2} + g^{2} + 2f.g = f^{2} + g^{2} – 2fg
⇒ fg = – fg. The equation can hold if f.g < 0 and f > g.
This can be simplified to f > 0, g < 0.
Answer the following questions on the basis of this method
Q. The complete solution of the equation x^{3} – x + 2 – x = x^{3} – 2 is
The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation f(x) + g(x) = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.
Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f^{2} + g^{2} + 2f.g = f^{2} + g^{2} – 2fg
⇒ fg = – fg. The equation can hold if f.g < 0 and f > g.
This can be simplified to f > 0, g < 0.
Answer the following questions on the basis of this method
Q. The complete solution set of the equation x^{2} – x + x + 3 = x^{2} – 2x – 3 is
The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation f(x) + g(x) = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.
Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f^{2} + g^{2} + 2f.g = f^{2} + g^{2} – 2fg
⇒ fg = – fg. The equation can hold if f.g < 0 and f > g.
This can be simplified to f > 0, g < 0.
Answer the following questions on the basis of this method
Q. The solution set belonging to (0, 2π) of the equation sin x – cos x = sin x + cosx is
Let f be a function defined so that every element of the codomain has at most two preimages and there is at least one element in the codomain which has exactly two preimages we shall call this function as “twoone” function. A twoone function is definitely a many one function but viceversa is not true. For example, y = e^{x} – 1 is a “twoone” function. y = x^{3} – x is a many one function but not a “twoone” function. In the light of above definition answer the following questions:
Q. In the following functions which one is a “twoone” function :
Let f be a function defined so that every element of the codomain has at most two preimages and there is at least one element in the codomain which has exactly two preimages we shall call this function as “twoone” function. A twoone function is definitely a many one function but viceversa is not true. For example, y = e^{x} – 1 is a “twoone” function. y = x^{3} – x is a many one function but not a “twoone” function. In the light of above definition answer the following questions:
Q. Let f(x) = {x} be the fractional part function. For what domain is the function “twoone”?
From graph it is clear that any horizontal line cut the graph at either one or two points between
Let f be a function defined so that every element of the codomain has at most two preimages and there is at least one element in the codomain which has exactly two preimages we shall call this function as “twoone” function. A twoone function is definitely a many one function but viceversa is not true. For example, y = e^{x} – 1 is a “twoone” function. y = x^{3} – x is a many one function but not a “twoone” function. In the light of above definition answer the following questions:
Q. The values of ‘a’ for which the function f(x) = x^{4} – ax^{2} is a “twoone” function are
(A)
(B)
(C)
(D)
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