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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A galvanometer of resistance 30 2 is connected to a battery of emf 2 V with 1970 Q resistance in series. A full scale deflection of 20 divisions is obtained in the galvanometer. To reduce the deflection to 10 divisions, the resistance in series required isA. `4030 Omega`B. `4000 Omega`C. `3970 Omega`D. `2000 Omega` |
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Answer» Correct Answer - C According to question, Net current through the galvanometer is given by `l=V/R_(eff)=2/(1970+30)=2/(2000)` = `10^-3` As we know, this current provides full scale deflection (i.e. 20 div). In order to limit the deflection to 10 divisions, the resistance needed to connect such that the current reduces can be obtained as `theta=(nlAB)/K ("i.e. "theta alpha l`) [Symbols have their usual meanings.] Righrarrow `theta_1/theta_2=l_1/l_2=2` `Rightarrow l_2=l_1/2=10^-3/2=5xx10^-4`A Again,`l=V/(R_(eff)+R_5)` `Rightarrow R_5=V/l-R_("eff")` `2/(5xx10^-4)-2000` `=4xx10^3-2000=2000 Omega` So, the resistance of `1970 Omega` is to be replaced by `1970+2000=3970 Omega.` |
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| 2. |
Identify the metal that froms colourless compounds.A. Iron (Z = 26)B. Chromium (Z = 24)C. Vanadium (Z = 23)D. Scandium (Z -21) |
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Answer» Correct Answer - D d-block elements show +3 as the most common oxidation state. For scandium (Z =21), electronic configuration is `d^1s^2`. After the removal of +3 electrons `(Sc^(3+))`, it acquires a stable configuration `(d^0)`. Hence, it forms colourless compounds. |
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| 3. |
The reaction takes place in two steps as (i) `NO_2Cl(g)overset(K_1)rarrNO_2(g)+Cl(g)` (ii) `NO_2Cl(g)+Cl(g)overset(K_1)rarrNO_2(g)+Cl_2(g)`A. `NO_2Cl(g)`B. `NO_2(g)`C. `Cl_2(g)`D. `Cl(g)` |
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Answer» Correct Answer - D CI (g) is the reaction intermediate involved in the formation of `NO_2(g)` and `Cl_2(g)`. Reaction intermediates are those which are formed within the reaction and has no presence in the products. CI (g) is involved in both the reactions. `NO_2Cl(g)overset(k_1)toNO_2(g)+Cl(g)` `NO_2Cl(g)+Cloverset(k_2)toNO_2(g)+Cl(g)` |
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| 4. |
Name the reagent that is used in leaching of gold.A. CarbonB. Sodium cyanideC. Carbon monoxideD. lodine |
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Answer» Correct Answer - B Leaching involves the treatment of the ore with a suitable reagent so as to make it soluble while impurities remain insoluble. Leaching of gold is done with the help of their dissolution in NaCNor KCN or Cu with the help of H or Fe scrap. `4Au(s)+8CN+O_2(g)+2H_2O rarr4[Au(CN)_2]^-+4OH^-` |
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| 5. |
Average rate of reaction for the following reaction, `2SO_(2)(g)+O_(2)(g)rarr2SO_(3)(g)` is written asA. `(Delta[SO_2])/(Deltat)`B. `(Delta[O_2])/(Deltat)`C. `1/2(Delta[SO_2])/(Deltat)`D. `(Delta[SO_3])/(Deltat)` |
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Answer» Correct Answer - B Average rate of reaction is defined as the speed with which the reactants are converted into products For reaction, `2SO_2(g)+O_2to2SO_3(g)-1/2(Delta[SO_2])/(Deltat)=(-Delta[O_2])/(Deltat)=+1/2(Delta[SO_3])/(Deltat)` So, average rate of reaction is written as `(-Delta[O_2])/(Deltat)` . |
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| 6. |
Identify an extensive property amongst the following:A. ViscosityB. Heat capacityC. DensityD. Surface tesion |
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Answer» Correct Answer - B Heat capacity is an extensive property. It is a property that changes when the size of the system changes whereas viscosity, density and surface tension are intensive properties. It is also termed as bulk property that does not depend on the system size. |
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| 7. |
How is sodium chromate converted into sodium dichromate in the manufacture of potassium dichromate from chromite ore?A. By the action of concentrated sulphuric acidB. By roasting with soda ashC. By the action of sodium hydroxideD. By the action of limestone |
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Answer» Correct Answer - A By the action of concentrated sulphuric acid, sodium chromate gets converted into sodium dichromate in the manufacture of potassium dichromate. Various steps involved are: Step I Concentration of the chromite ore by Wilfley method. Step II Conversion of the concentrated chromite ore into `Na_2Cro_4`. Concentrated chromite ore is fused with molten `Na_2CO_3` in the presence of air when `Na_2CrO_4` and `Fe_2O_3` are formed. Step III Conversion of `Na_2CrO_4` into `Na_2Cr_2O_7`. Aqueous solution of `Na_2CrO_4` is acidified to `Na_2Cr_2O_7` `Na_2CrO_4+H_2underset("Dil")SO_4toNa_2Cr_2O_7+H_2O` |
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| 8. |
Identify the compound amongst the following of which 0.1 M aqueous solution has highest boiling point.A. GlucoseB. Sodium chlorideC. Calcium chlorideD. Ferric chloride |
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Answer» Correct Answer - D As all the compounds have same concentration i.e, 0.1M. Thus, the compound that will break into the most parts has highest boiling point. Glucose (! Part, covalent, does not ionise) Sodium chloride (1NA, 1Cl, ionises) Ferric Chloride (1Fe, 3Cl, ionises) |
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| 9. |
In dry cell, what acts as a negative electrode?A. ZincB. GraphiteC. Ammonium chlorideD. Manganese dioxide |
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Answer» Correct Answer - A In a dry cell, zinc acts as a negative electrode. In this cell, the outer zinc container is the negatively charged terminal. At anode (negative terminal) `Zn (s)to Zn^(2+) (aq)+ 2e^-` At cathode (positive terminal) `2MnO_2+2e^-+2NH_4Cl(aq)toMn_2O_3(s)+2NH_3(aq)+H_2O(l)+2Cl^-` |
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| 10. |
Which of the following is not an analgesic?A. OfloxacinB. PenicillinC. AminoglycosidesD. Paracetamol |
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Answer» Correct Answer - D Paracetamol is not an analgesic. It is an antipyretic. The chemical substances which are used to bring down body temperature during high fever are called antipyretics and the medicines which are used to reduce pain are known as analgesics. |
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| 11. |
Which halide of magnesium has highest ionic character?A. ChlorideB. BromideC. lodideD. Fluoride |
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Answer» Correct Answer - D Magnesium fluoride (MgF) has the highest ionic character. In ionic bonding, there is large electronegativity difference between two atoms, causing the bonding to be more polar have maximum ionic character `Mgl_2 lt MgBr_2 lt MgCl_2 lt MgF_2`. |
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| 12. |
Which of the following compound when treated with dibenzyl cadmium yields benzyl methyl ketone?A. AcetoneB. AcetaldehydeC. Acetic acidD. Acetyl chloride |
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Answer» Correct Answer - D Acetyl chloride when treated with dibenzyl cadmium yields benzyl methyl ketone. `2CH_(3)-underset(O)underset(||)(C)-Cl+Ph_(2)Cd to 2CH_(3)-underset(O)underset(||)(C)-Ph+CdCl_(2)` |
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| 13. |
If the end correction of an open pipe is 0.8 cm, then the inner radius of that pipe will beA. `1/3`cmB. `2/3` cmC. `3/2` cmD. 0.2cm |
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Answer» Correct Answer - C For an open organ pipe, the relation between end correction and inner radius of the organ pipe is given by `Deltal=1.2xxr` [`Deltal` end correction, r= inner radius] So, `r=(Deltal)/1.2=(0.8)/1.2` `[because Deltal = 0.8]` `2/3cm` |
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| 14. |
A progressive wave is represented by `y = 12 sin (5t - 4x)` cm. On this wave, how far away are the two points having phase difference of `90^circ` ?A. `pi/2cm`B. `pi/4cm`C. `pi/8cm`D. `pi/16cm` |
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Answer» Correct Answer - C According to question, the progressive wave is represented by y = `12 sin (5t – 4x)` cm Comparing this equation with standard equation of progressive wave, `y=sin(omegat-kx)` So we have A=12 `omega =5` Rightarrow`k=4` Here, `(omegat-kx)` is phase difference = `pi/2` `therefore 5t-4t=pi/3` When `t=0,4x=pi/2cm` `therefore x=pi/8`cm |
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| 15. |
A ray of light travelling through rarer medium is incident at very small angle i on a glass slab and after refraction its velocity is reduced by 20%. The angle of deviationA. `i/8`B. `i/5`C. `i/2`D. `(4i)/5` |
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Answer» Correct Answer - B According to question, at glass-air interface velocity is reduced by 20% of the velocity of light. So, deviation =20% of I = `(20xx i)/(100)=(i)/(5)` |
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| 16. |
Light of wavelength A which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with same velocity, then stopping potential willA. increaseB. decreaseC. be zeroD. become exactly half |
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Answer» Correct Answer - A According to photoelectric equation, `(hc)/lambda = phi+E` Symbols have their usual meaning E =kinetic energy If E is constant, then `1/lambda alpha phi=` If we decrease the wavelength lambda, then stopping potential `phi` will increase such that `(hc)/lambda-phi` = constant |
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| 17. |
In an oscillator, for sustained oscillations, Barkhausen criterion is `Abeta` equal to (A = voltage gain without feedback and `beta` = feedback factor)A. zeraB. `1/2`C. 1D. 2 |
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Answer» Correct Answer - C Barkhausen criterion states that if A is the gain of the amplifying element in the circuit and B is the transfer function of the feedback path, then condition of sustained oscillation is given by |`beta A|=1` So, option (c) is correct. |
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| 18. |
Brown ring test is used for detection of which radical?A. FerrousB. NitriteC. NitrateD. Ferric |
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Answer» Correct Answer - C Brown ring test is used for the detection of nitrate radical. `2NaNO_3=2H_2SO_4to2NaHSO_4+2HNO_3` `2HNO_3+6FeSO_4+3H_2SO_4to3Fe_2(SO_4)_3+2NO+4H_2O` `[Fe(H_2O)_6]SO_4+NOto[Fe(H_2O)_5NO]^(2+)SO_4^(2-)+H_2O` (Brown ring) |
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| 19. |
The criterion for a spontaneous process isA. `DeltaGgt0`B. `DeltaGlt0`C. `DeltaG=0`D. `DeltaS_(Total)lt0` |
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Answer» Correct Answer - B The criterion for a spontaneous process is `DeltaG lt 0` Gibbs free energy is the maximum amount of energy available to a system during a process, that can be converted into useful work. `DeltaG=DeltaH-TDeltaS` It is called Gibbs-Helmholtz equation and useful in predicting the spontaneity of process. If `DeltaG=-ve`, the process is spontaneous. If `DeltaG= +ve`, the process is non-spontaneous. If `DeltaG = zero`, the process is in equilibrium. |
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| 20. |
If matrix `A=[(1,2),(4,3)]`, such that `AX=l`, then `X` is equal toA. `(1)/(5)]{:(,1,3),(,2,-1):}]`B. `(1)/(5)[(,4,2),(,4,-1):}]`C. `(1)/(5)[{:(,-3,2),(,4,-1):}]`D. `(1)/(5)[{:(,-1,2),(,-1,4):}]` |
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Answer» Correct Answer - C Given `A=[{:(,1,2),(,4,3):}]` and AX=I `Rightarrow X=A^(-1)l` `Rightarrow X=A^(-1)` Now, `A^(-1) =(1)/(|A|) [{:(,3,-2),(,-4,1):}]` `=(1)/(3-8) [{:(,3,-2),(,-4,1):}]` `=(1)/(-5) [{:(,3,-2),(,-4,1):}]` `=(1)/(5)[{:(,-3,2),(,4,-1):}]` |
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| 21. |
The rate constant and half - life of a first order reaction are related to each other as _____.A. `t_(1//2)=(0.693)/k`B. `t_(1//2)=0.693`C. `k=0.693t_(1//2)`D. `kt_(1//2)=1/0.693` |
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Answer» Correct Answer - A For first order reaction, `k=(2.303)/t"log "[[Ag]]/[[A]]` At `t=t_(1//2),[A]=[A_0]//2` `k=(2.303)/t_(1//2)(log) [[A_0]]/[[A_0]],k=(2.303)/t_(1//2)log2` `k=(0.693)/t_(1//2)or t_(1//2)=(0.693)/k` |
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| 22. |
Mathematical equation of first law of thermodynamics for isochoric process isA. `DeltaU=qv`B. `-Delta=qv`C. `q=-W`D. `DeltaU=W` |
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Answer» Correct Answer - A For an isochoric process, `DeltaV=0` In this process, the first law of thermodynamics `q=DeltaU-W` `q=DeltaU-PDeltaV` As `DeltaV=0` `qV=DeltaU` |
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| 23. |
The acute angle between the line `r=(hati+2hatj+hatk)+lambda(hati+hatj+hatk)` and the plane `r.(2hati-hatj+hatk)=5`A. `cos^(-1)((sqrt2)/(3))`B. `sin^(-1)((sqrt2)/(3))`C. `tan^(-1)((sqrt2)/(3))`D. `sin^(-1)((sqrt2)/(3))` |
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Answer» Correct Answer - B We know that, the angle between the line `r=a+lambdab` and the planer r.n=d is `sin theta =(n.b)/(|n||b|)` Given equation of line is `r=(hati+2hatj+hatk)+lambda(hati+hatj+hatk)` and plane is `r.(2hati-hatj+hatk)=5` Here, `b=hati+hatj+hatk and n=2hati-hatj+hatk` `therefore sin theta=((2hati-hatj+hatk).(hati+hatj+hatk))/(|2hati-hatj+hatk|hati+hatj+hatk))` `=(2-1+1)/(sqrt(2^2+(-1)^(2)+(1)^(2)sqrt(1^(2)+1^(2)+1^(2))` `(2)/(sqrt(4+1+1)sqrt(1+1+1))=(2)/(sqrt6 sqrt3))=(2)/(3sqrt2))` `Rightarrow sin theta=(sqrt2)/(3)` `Rightarrow theta=sin^(-1) ((sqrt2)/(3))` |
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| 24. |
5.0 g of sodium hydroxide (molar mass 40 g mol) is dissolved in little quantity of water and the solution is diluted upto 100 mL. What is the molarity of the resulting solution?A. `0.1" mol dm"^-3`B. `1.0" mol dm"^-3`C. `0.125" mol dm"^-3`D. ` 1.25" mol dm"^-3` |
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Answer» Correct Answer - D Given, `W ("mass of solute") = 5.0g` m (molar mass of solute)= `40" g mol"^-1` Volume 100 mL Molarity is given as, `M=("Moles of solute")/("Volume of solution in L")` Acetyl chloride when treated with dibenzyl cadmium yields benzyl methyl ketone Here, Ph=Phenyl group |
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| 25. |
In the cell represented by `Pb(s)|Pb^(2+)(1 M)||Ag^(+)(1M)|Ag(s)` the reducing agent isA. `Pb`B. `Pb^(2+)`C. AgD. `Ag^+` |
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Answer» Correct Answer - A In the cell represented by `Pb(s) |Pb^(2+) (1M)| |Ag^+(1M) | Ag(s)`, the reducing agent is Pb because it readily gets oxidised to `Pb^(2+)` and helps in the reduction of `Ag^++etoAg(s)` |
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| 26. |
If `G(barg), H(barh)` and `P(barp)` are centroid, orthocenter and circumcenter of a triangle and `xbarp + ybarh + zbarg =0` then `(x, y, z)=`A. 1,1,-2B. 2,1,-3C. 1,3,-4D. 2,3,-5 |
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Answer» Correct Answer - B We know that, orthocentre, centroid and circumcentre of a triangle are collinear and centroid divides orthocentre and circumcentre in the ratio 2 : 1 By using internally division `(2p+1h)/(2+1)=g` `Rightarrow 2p+h-3g=0` But it is given xp+yh+zg=0 `therefore x=2, y=1 and z=-3` |
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| 27. |
If a random variable waiting time in minutes for bus and probability density function of x is given by `f(x)={{:(1/5","0lexle5),(0",""otherwise"):}` Then probability of waiting time not more than 4 minutes is equal toA. 0.3B. 0.8C. 0.2D. 0.5 |
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Answer» Correct Answer - B Given, `f(x)={{:(,(1)/(5), 0 le x le 4),(,0,"otherwise"):}` `therefore P(0 le x le 4)=underset(0)overset(4)f(x)dx=underset(0)overset(1) dx=(1)/(5)[x]_(0)^(4)` `=(1)/(5)(4-0)=(4)/(5)=0.8` |
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| 28. |
If `y = e^(m) sin^(-1) x and (1-x^(2)) ((dy)/(dx))^(2) = At^(2)` , then A is equal toA. mB. `-m`C. `m^(2)`D. `-m^(2)` |
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Answer» Correct Answer - C Given `y=e^(m^(sin^(-1)))x ......(i)` On differentiating both sides w.r.t.x. we get `(dy)/(dx)=e^(m^(sin(-1))) (d)/(dx)(m sin^(-1)x)` `Rightarrow (dy)/(dx)=e^(m^(sin^(-1))) (mxx(1)/(sqrt(1-x^(2))))` `Rightarrow sqrt(1-x)^(2) (dy)/(dx)=my " "("from Eq. (i)")` On squaring both sides, we get `(1-x^(2)) ((dy)/(dx))^(2)=m^(2)y^(2)` But it is given `(t-x^(2)) ((dy)/(dx))^(2)=Ay^2` `A=m^(2)` |
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| 29. |
`(tan^(-1)(sqrt3)-sec^(-1)(-2))/(cosec^(-1)(-sqrt2)+cos^(-1)(-(1)/(2)))` is equal toA. `(4)/(5)`B. `-(4)/(5)`C. `(3)/(5)`D. 0 |
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Answer» Correct Answer - B `(tan^(-1)(sqrt3)-(pi-sec^(-1)(2)))/(-cosec^(-1)(sqrt2)+pi-cos^(-1)((1)/(2)))` `=((pi)/(3)-(pi-(pi)/(3))/(-(pi)/(4)+pi-(pi)/(4))=((2pi)/(3)-x))/(pi-(pi)/(4)-(pi)/(3))` `=(-(pi)/(3))/((12pi-3pi-4pi)/(12))=(-(pi)/(3))/((5pi)/(12))=-(4)/(5)` |
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| 30. |
For what value of k, the function defined by `f(x)={((log(1+2x)sin x^(@))/(x^(2))",", "for "x ne 0),(k",","for "x =0):}` is continuous at x = 0 ?A. 2B. `(1)/(2)`C. `(pi)/(90)`D. `(90)/(pi)` |
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Answer» Correct Answer - C Given `p(x)={{:(,(log(1+2x)sinx^(@))/(x^(2)),"for x "ne 0),(,k, "for x=0"):}` `={{:(,(log(1+2x)sin""(pix)/(180))/(x^(2)),"for x"ne 0),(,k,"for x=0"):}` Since, f(x) is continuous at x=0 `therefore LHL=f(0)` Now, `LHL=underset(x to 0^(-))lim f(x)=underset(h to 0)lim f(0-h)` `=underset(h to 0)lim (log(1+2(0-h)sin""(pi)/(180^(@))(0-h))` `=underset(h to 0) (log (1+2(0-h)sin""(pi)/(180^(@))(0-h))/((0-h)^(2))` `=underset(h to 0) (log (1-2h){-sin""(pih)/(180)})/(h^(2))` `=underset(h to 0)lim (-2) (log (1-2h))/(-2h))=x^(-) underset(h to 0)lim (sin""(pih)/(180))/((pih)/(180))xx(pi)/(180)` `=(-2) xx(-1) xx 1 xx (pi)/(180) " "[therefore underset(x to 0)lim log ""(1+x)/(x)=1 and underset(x to 0)lim (sin x)/(x)=1]` `=(pi)/(90)` |
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| 31. |
If `A=[(1,1,0),(2,1,5),(1,2,1)]` then `a_(11)A_(21)+a_(12)A_(22)+a_(13)A_(23)` is equal toA. 1B. 0C. `-1`D. 2 |
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Answer» Correct Answer - B Given `A=[{:(,1,1,0),(,2,1,5),(,1,2,1):}]` ltBRgt We know that, the sum of product of element other than the corresponding cofactor is zero. `therefore a_(11)A_(21)+a_(12)A_(22)+a_(13)A_(23)=0` |
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| 32. |
Let `X~B(n,p)`, if E(X)=5,Var(X)`=2.5` then `P(X lt 1)` is equal toA. `((1)/(2))^(11)`B. `((1)/(2))^(10)`C. `((1)/(2)))^(6)`D. `((1)/(2))^(9)` |
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Answer» Correct Answer - B Given mean E(x)=5 and variance, Var(X)=2.5 `therefore np=5 and npq=2.5` `Rightarrow 5q=2.5=q=(1)/(2)` Also, p+q=1 `therefore p=1-(1)/(2)=(1)/(2)` `therefore np=5` `Rightarrow np=5` `Rightarrow n xx(1)/(2)=5 " "[therefore p=(1)/(2)]` `Rightarrow n=10` `p(X lt 1)==p(X=0)=""^(n)C_(r) p^(r) q^(n-r)` `=""^(10)C_(0) ((1)/(2))^(0) ((1)/(2))^(10-0)` `=1xx 1xx((1)/(2))^(10)=((1)/(2))^(10)` |
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| 33. |
`int(dx)/(sqrt(8+2x-x^(2)))`A. `(1)/(3)sin^(-1) ((x-1)/(3))+c`B. `sin^(-1)((x+1)/(3))+C`C. `(1)/(3)sin^(-1) ((x+1)/(3))+c`D. `sin^(-1)((x-1)/(3))+c` |
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Answer» Correct Answer - D `int (dx)/(sqrt(8+2x-x^(2))=int (dx)/(sqrt((8-(x^(2)-2x+1)+1))` `int (dx)/(sqrt(3^(2)-(x-1)^(2))=sin^(-1) ((x-1)/(3))+c` |
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| 34. |
The general solution of the equation `tan^(2)x=1` isA. `n pi+(pi)/(4)`B. `n pi-(pi)/(4)`C. `n pi (pi)/(4)`D. `2n pi pm (pi)/(4)` |
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Answer» Correct Answer - C Given `tan^(2)x=1` `Rightarrow tan^(2)x=1^(2)` `Rightarrow tan^(2)x=tan^(2)""(pi)/(4)` `Rightarrow x=n pi pm (pi)/(4)` |
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| 35. |
`int((4e^x-25)/(2e^x-5))dx=A x+B log/(2e^x)-5/(+c)` thenA. A=5 and B=3B. A=5 and B=-3C. A=-5 and B=3D. A=-5 and B=-3 |
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Answer» Correct Answer - B Let `t=int ((4e^(x)-25)/(2e^(x)-5))dx` `=int (4e^(x))/(2e^(x)-5))dx-int (25)/(2e^(x)-5))dx` `=4int (e^(x))/(2e^(x)-5))dx=25 int (e^(-x))/(2-5e^(-x))=dx` Put `2e^(x)-5=u and 2-5e^(-x)=v` `Rightarrow 2e^(-x)dx=dv` `and 5e^(-x)dx=dv` `Rightarrow e^(x)dx=(du)/(2) and e^(-x)dx=(dv)/(5)` `therefore I=4 int (du)/(2u)-25 int (du)/(5v)` `=2log u-5log v+c` `=2 log (2e^(x)-5)-5log(2-5e^(-x))+c` `=2log(2e^(x)-5)-5log((2e^(x)-5)/(e^(x)))` `=2log (2e^(x)-5)-5log (2-5e^(-x)+c` `=-3log (2e^(x)-5)+5x+c` `Rightarrow I=5x-3log (2e^(x)-5)+c` But it is given I=Ax+Blog `(2e^(x)-5)+c` `therefore A=5 and B=-3` |
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| 36. |
If `2 tan^-1 (cosx)=tan^-1 (2 cosecx)` then `sinx + cosx=`A. `2sqrt2`B. `sqrt2`C. `(1)/(sqrt2)`D. `(1)/(2)` |
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Answer» Correct Answer - B Given `2tan^(-1)(cos x)=tan^(-1)(2cosec x)` `Rightarrow tan^(-1) (2cos x)/(1-cos^(2)x)=tan^(1)((2)/(sin x))` `Rightarrow (2cos x)/(1-cos^2x)=(2)/(sin x) Rightarrow (cos x)/(sin^(2)x)=(1)/(sin x)` `Rightarrow (cos x)/(sin x)=1 [therefore sin x ne 0]` `Rightarrow tan x=1 Rightarrow x= (pi)/(4)` Now, `sin x+cos x=sin"" (pi)/(4)+cos ""(pi)/(4)` `=(1)/(sqrt2)+(1)/(sqrt2)=sqrt2` |
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| 37. |
Derivative of ` tan^(-1)((x)/(sqrt( 1 - x^(2))))` with respect to ` sin^(-1) (3x - 4x^(3)) ` isA. `(1)/(sqrt(1-x^(2))`B. `(3)/(sqrt(1-x^(2)))`C. 3D. `(1)/(3)` |
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Answer» Correct Answer - D Let `u=tan^(-1) ((x)/(sqrt(1-x^(2))` and `v=sin^(-1)(3x-4x^(3))` put x= `sin theta`, then `u=tan^(-1) ((sin theta)/(sqrt(1-sin^(2)theta))` and `v=sin^(-1)(3 sin theta-4sin^3 theta)` `Rightarrow u=tan^(-1) ((sin theta)/(cos theta))` and `v=sin^(-1) (sin 3 theta)` `Rightarrow u=tan^(-1) (tan theta)` and `v=sin^(-1)(sin 3 theta)` `Rightarrow u=theta and v=3 theta` `Rightarrow u=sin^(-1) x and v=3 sin^(-1) x`. On differentiating both sides w.r.t.x. we get `(du)/(dx)=(1)/(sqrt(1-x^(2)) and (dv)/(dx)=3 xx (1)/(sqrt(1-x^(2))` `therefore therefore (du)/(dv)=((du)/(dx))/((dv)/(dx))= ((1)/(sqrt(1-x^(2))/((3)/(sqrt(1-x^(2)))=(1)/(3)` |
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| 38. |
`((int (x)/(2))/(-(x)/(2)))log((2-sinx)/(2+sin x))` dx is equal toA. 1B. 3C. 2D. 0 |
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Answer» Correct Answer - D We have, `i=(underset(-x)overset((pi)/(2))int)/(2) log ((2-sin x)/(2+sin x))dx` `"Let " f(x)=log ((2-sinx)/(2+sin x))` Then, f(-x)=`log ((2-sin(-x))/(2+sin(-x)))` `=log ((2+sin x)/(2-sinx))=log ((2-sinx)/(2+sin x))^(-1)` `=-log ((2-sin x)/(2+sin x))=-f(x)` Then, f(x) is an odd function. `therefore (underset(pi)overset((pi)/(2))int (2) f(x)dx=0` `[therefore ` If f(x) is an odd function, then `underset(-a)overset(a) f(x)dx=0` |
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| 39. |
Primary nitroalkanes are obtained in good yield by oxidising aldoximes with the help ofA. trifluoroperoxy acetic acidB. acidified potassium permanganateC. concentrated nitric acidD. potassium dichromate and dilute sulphuric acid |
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Answer» Correct Answer - A Primary nitroalkanes are obtained in good yield by oxidising aldoximes with the help of trifluoroperoxy acetic acid. `underset("Aldoxime")(R CH = NOH) underset(CF_(3)COOOH)overset([O])rarr RCH = overset(O^(-))overset(|)(N^(+))-OH hArr underset((1^(@) "nitroalkane"))(R CH_(2)NO_(2))` |
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| 40. |
The joint equation of bisectors of angles between lines `x=5` and `y=3` isA. `(x-5)(y-3)=0`B. `x^2-y^(2)-10x+6y+16=0`C. `x^2-y^(2)-10x+6y+16=0`D. `xy=0` |
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Answer» Correct Answer - B The equation of the bisector of the angle between the lines `(x-5) and (y-3)` is `(((x-5))/(sqrtt^(2)))= pm ((y-3)/(sqrtt^(2))) Rightarrow (x-5)/(1)= pm (y-3)/(1)` `Rightarrow x-5=+ (y-3) and x-5=-(y-3)` `Rightarrow (x-y-2)=0 and (x+y-8)=0` `therefore ` Combined equation of bisector of angle between the lines is `(x-y-2)(x+y-8)=0` `Rightarrow x^(2)+xy-8x-xy-y^(2)+8y-2x-2y-16=0` `Rightarrow x^(2)-y^(2)-10x+6y+16=0` |
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| 41. |
Derivative of ` log (sec theta + tan theta )` with respect top ` sec theta ` at ` theta = (pi)/(4) ` is |
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Answer» Correct Answer - B Let u=log `(sec theta+tan theta) and v=sec theta`. On differentiating both sides w.r.t.x. `theta` we get `(du)/(dtheta)=(1)/((sec theta+tan theta))(sec theta tan theta+sec^(2)theta) and (dv)/(d theta)=sec theta tan theta` `therefore (du)/(dv)=((du)/(d theta))/((dv)/(d theta))` `=((sec theta (tan theta+sec theta))/((sec theta+ tan theta)xx sec theta tan theta))=cot theta` `Rightarrow (du)/(dv)_((theta=(pi)/(4)))=cot""(pi)/(4)=1` |
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| 42. |
Alternating current of peak value `((2)/(pi))` ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak e.m.f. induced in secondary coil is (Frequency of AC= 50 Hz)A. 100VB. 200VC. 300VD. 400V |
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Answer» Correct Answer - B According to question, peak value of current `l_0 =sqrt2xxl_(rms)=2/piA` Coefficient of mutual inductance 1H As we know, Induced emf in secondary coil is given by `E_s=M.(dl)/dt` [`Where,l=l_0sinomegat`] `E_s=Momegal_0cos(omegat)` =`1xx2pixx50xx2/picos(2pixx50xxt)` `because omega=2pin` For t= 0 we have `E_s 4xx50 =200V` |
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| 43. |
Which of the following equation does not represent a pair of lines ?A. `x^(2)-x=0`B. `xy-x=0`C. `y^2-x+1=0`D. `xy+x+y+1=0` |
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Answer» Correct Answer - C `x^(2)-x=0` `Rightarrow x(x-1)=0` `Rightarrow x=0 and x=1` Which represent a pair of lines. `xy-x=0` `Rightarrow x(y-1)=0` `Rightarrow x=0, y=1` Which represent a pair of lines. `y^(2)-x+1=0` `Rightarrow y^2=(x-1)` Which represent a parabola. Thus, it does not represent a pair of lines. `xy+x+y+1=0` `Rightarrow x(y+1)+(y+1)=0` `Rightarrow (y+1)(x+1)=0` `Rightarrow u=-1 and x=-1` Which represent a pair of lines. |
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| 44. |
Principal solutions of the equation `sin 2x + cos 2x = 0`, where `pi < x < 2pi `A. `(7pi)/(8),(11pi)/(8)`B. `(9pi)/(8),(13pi)/(8)`C. `(11pi)/(8),(15pi)/(8)`D. `(15pi)/(8),(19pi)/(8)` |
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Answer» Correct Answer - C Given equation in sin 2x+cos2x=0 `Rightarrow sin 2x=-cos 2x` `Rightarrow tan 2x=-1" "[therefore pi t x lt 2pi Rightarrow 2pi lt 2x lt 4pi]` `Rightarrow 2x=2pi +(3pi)/(4), 2pi+((3pi)/(2)+(pi)/(4))` `Rightarrow 2x=(11x)/(8),(15pi)/(4)` `Rightarrow x=(11pi)/(8), (15pi)/(8)` |
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| 45. |
If `A=[(2,2),(-3,2)], B=[(0,-1),(1,0)]` then `(B^(-1)A^(-1))^(-1)` is equal toA. `[{:(,2,-2),(,2,3):}]`B. `[{:(,2,2),(,-2,3):}]`C. `[{:(,2,-3),(,2,2):}]`D. `[{:(,1,-1),(,-2,3):}]` |
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Answer» Correct Answer - A Given, `A-[{:(,2,2),(,-3,2):}], B=[{:(,0,-1),(,1,0):}]` `therefore (B^(-1)A^(-1))=(A^(-1))^-1(B^(-1))^(-1) [therefore (AB)^(-1)=B^(-1)A^(-1)]` `AB=" "[therefore (A^(-1)^(-1)=A^(-1))` `=[{:(,2,2),(,-3,2):}] [{:(,0,-1),(,1,0):}]` `=[{:(,0+2,-2+0),(,0+2,3+0):}]=[{:(,2,-2),(,2,3):}]` |
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| 46. |
The bob of a simple pendulum performs SHM with period T in air and with period `T_(1)` in water. Relation between T and `T_1` is (neglect friction due to water, density of the material of the bob is = `9/8xx 10^3 kgm^3`, density of water = `1gcc^-1`)A. `T_1 = 3T`B. `T_1 = 2T`C. `T_1 = T`D. `T_1 = T/2` |
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Answer» Correct Answer - A Time period of simple pendulum in water is given by `T=2pisqrt(l/g_(eff))` [Symbols have their usual meanings and `g_(eff)` = acceleration due to gravity in water] As we khow, `g_(eff)=g((sigma-rho)/sigma)` [Where,`sigma` = density of bob, p = density of water] `9.8((9/8xx10^3-10^3)/(9/8xx10^3))` `9.8((9/8-1)/(9/8))=9.8(1/8xx8/9)` `(9.8)/9` So, `T_1=2pisqrt((lxx9)/9.8) Rightarrow T_1=3T` [therefore Time period of simple pendulum in air, `T_1=2pisqrt((l)/9.8)` ] |
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| 47. |
If line joining points A and B having position vectors `6 bar a-4 bar b+4 bar c` and `-4 bar c` respectively, and the line joining the points C and D having position vectors `-bar a-2 bar b-3 bar c` and `bar a+2 bar b -5 bar c` intersect, then their point of intersection is(A) B(B) C(C) D(D) AA. BB. CC. DD. A |
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Answer» Correct Answer - A Coordinate of points A and B are (6, -4, 4) and ( 0, 0, -4) and coordinate of points C and D are (-1, -2, -3) and (1, 2, -5). Now, equation of line passing through (0, 0, -4) and (6, -4,4) is `(x-0)/(6)=(y-0)/(-4) =(z+4)/(4+4)=k" "["say"]` `Rightarrow x=6k, y=-4k` and `z=8k-4....(i)` Again, equation of line passing through (-1,-2,-3) and (1,2,-5) is `(x+1)/(1+1)=(y+2)/(2+2)=(z+3)/(-5+3)` `Rightarrow (x+1)/(2)=(y+2)/(4)=(3+3)/(-2).....(ii)` Since, two lines are intersect, therefore point (6k, -4k, Bk -4) satisfy Eq. (ii), we get `(6k+1)/(2)=(-4k+2)/(4)=(8k-4+3)/(-2)` `Rightarrow 6k+1=-2k+1=-(8k-1)` `therefore 6k+1=-2k+1` `Rightarrow 8k=0` `Rightarrow k=0` `therefore x=6 xx 0, y=-4 xx 0` and `z=8xx0-4` `Rightarrow x=0, y=0 and z=-4 which is equal to the B coordinate. |
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| 48. |
Magnetic flux passing through a coil is initially `4xx 10^-4` Wb. It reduces to 10% of its original value in t second. If the emf induced is 0.72 mV then t in second isA. `0.3`B. `0.4`C. `0.5`D. `0.6` |
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Answer» Correct Answer - C According to question, Initial flux `(phi_1) = 4 xx10^-4` Wb Final flux `(phi_2)=(4xx10^-4xx10)/100` =`4xx10^-5` Wb Emf induced is given by `e=|-(dphi)/dt| Rightarrow(dphi)/dt` Rightarrow 0.72xx10^-3=(4xx10^-4xx10^-5)/t` Rightarrow `t=(4xx10^-5xx9)/(72xx10^-5)=1/2=0.5`s |
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| 49. |
A mass m, connected to a horizontal spring performs SHM with amplitude A. While mass m, is passing through mean position, another mass m, is placed on it so that both the masses move together with amplitude A,. The ratio ofA. `[m_1/(m_1+m_2)]^(1/2)`B. `[(m_1+m_2)/m_1]^(1/2)`C. `[m_2/(m_1+m_2)]^(1/2)`D. `[(m_1+m_2)/m_2]^(1/2)` |
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Answer» Correct Answer - A Potential energy of oscillating mass is given by `PE=1/2kx^2` When only one block is oscillating at x=A `E_("max")=1/2KA^2` and at mean position x=0. So, E=0 `therefore A alpha 1/sqrtm` When mass `m_(2)` is placed on top of the mass m. Then, total mass is `(m_1 + m_2)` and E=0 at this point, as x = 0. lt brgt When they reaches at x = A, When they reaches at x = A, then `A_1prop 1/sqrt(m_1+m_2)` Dividing Eq. (ii) by Eq. (i), `A_1/A=((m_1)/(m_1+m_2))^(1/2)` |
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| 50. |
A disc of radius R and thickness has moment of inertia / about an axis passing through its centre and perpendicular to its plane. Disc is melted and recast into a solid sphere. The moment of inertia of a sphere about its diameter isA. `l/5`B. `l/6`C. `l/32`D. `l/64` |
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Answer» Correct Answer - A According to question, Moment of inertia of disc is given by `l=(MR^2)/2` [Symbals have their usual meanings.] When the disc is remoulded into solid sphere, then volume remains same. i.e,Volume of disc =Valume of solid sphere i.e `piR^2xxR/6=4/3pir^3` `Rightarrow r^3=R^3/8 Rightarrow r=R/2` Now, moment of inertia of solid sphere is given by `2/5mr^2` `=2/5xxmxxR^2/4=(mR^2)/10` `l/5` [l = moment of inertia of disc] |
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