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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 15051. |
5 pencils and 6 erasers cost is equal to 33 rupees and the cost a pencil and an eraser is 6 rupees, find out the cost of eraser.1. 12. 23. 34. 4 |
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Answer» Correct Answer - Option 3 : 3 Given : The cost of 5 pencils and 6 erasers is equal to 33 rupees The cost of a pencil and an eraser is equal to 6 rupees Calculations : Let the cost of a pencil be P Let the cost of an eraser be E According to the question 5P + 6E = 33 ....(1) P + E = 6 ....(2) Multiply equation (2) by 5 and solving equation (1) and (2) P = 3 and E = 3 ∴ The cost of the eraser is 3 rupees |
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| 15052. |
Energy levels `A, B, C` of a certain atom corresponding to increasing values of energy i.e., `E_(A) lt E_(B) lt E_(C)`. If `lambda_(1), lambda_(2), lambda_(3)` are the wavelengths of radiations correspnding to the transitions `C` to `B, B` to `A` and `C` to `A` respectively, which o fthe following statements is correct? A. `lambda_(3) = lambda_(1)+lambda_(2)`B. `lambda_(3)=(lambda_(2)lambda_(1))/(lambda_(1)+lambda_(2))`C. `lambda_(1)+lambda_(2)+lambda_(3) =0`D. `lambda_(3)^(2) = lambda_(1)^(2) +lambda_(2)^(2)` |
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Answer» Correct Answer - B `E_(C)-E_(b) = (hc)/(lambda_(1))` `E_(C) - E_(A) = (hc)/(lambda_(3))` `E_(B) - E_(A) = (hc)/(lambda_(3))` Solving the equations and get the answer. |
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| 15053. |
Two bodies A and B are at position `(2, 3)m` amd `(5, 7)m` respectively. These two are moving with constant velocity `3 ms^(-1)` along `-ve` y axis and `4ms^(-1)` along `+ve` x axis respectively. Minimum separation between them during their motion isA. `5 m`B. `3 m`C. `4 m`D. `10 m` |
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Answer» Correct Answer - A Since `bar(V)_(AB)` and `bar(R)_(BA)` are antiparrallel to each other initial separation is minimum at initial state |
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| 15054. |
Hydrogen `(_(1)H^(1))` Deuterium `(_(1)H^(2))` singly omised helium `(_(1)He^(1))` and doubly ionised lithium `(_(1)Li^(6))^(++)` all have one electron around the nucleus Consider an electron transition from `n = 2 to n = 1` if the wavelength of emitted radiartion are `lambda_(1),lambda_(2), lambda_(3),and lambda_(4)`, repectivelly then approximetely which one of the following is correct ?A. `lambda_(1)=lambda_(2)=4lambda_(3)=9lambda_(4)`B. `4lambda_(1)=2lambda_(2)=2lambda_(3)=lambda_(4)`C. `lambda_(1)=2lambda_(2)=2sqrt(2)lambda_(3)=3sqrt(2)lambda_(4)`D. `lambda_(1)=lambda_(2)=2lambda_(3)=3lambda_(4)` |
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Answer» Correct Answer - A `(1)/(lambda)=R((1)/(1)-(1)/(4))z^(2)` `lambda_(1):lambda_(2):lambda_(3):lambda_(4)=1:1:(1)/(4):(1)/(9)` `lambda_(1)=lambda_(2)=4lambda_(3)=9lambda_(4)` |
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| 15055. |
The transfer ration of a transistor is `50`. The input resistance of the transistor when used in the common -emitter configuration is `1 kOmega`. The peak value for an `A.C.` input voltage of `0.01 V` peak isA. `0.25 muA`B. `0.01 muA`C. `500 muA`D. `100 muA` |
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Answer» Correct Answer - c Input current `=(V_(i))/(R_(i))=(0.01)/(10^(3))= 10^(-5)A` Output collector current `=Bxx "input current"` `= 50xx10^(-5)` `=500A` |
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| 15056. |
A common emitter amplifier has a voltage gain of `50`, an input impedence of `100 Omega` and an output impedence of `200 Omega`. The power gain of the of the amplifier isA. `500`B. `1000`C. `1250`D. `100` |
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Answer» Correct Answer - C `AC` power gain `= ("Change in output power")/("Change in input power")` `= (DeltaV_(c) xx Deltai_(c))/(DeltaV_(i) xx Deltai_(b)) = ((DeltaV_(c))/(DeltaV_(i))) xx ((Deltai_(c))/(Deltai_(b)))` `= A_(V) xx beta_(AC)` where `A_(V)` is voltage gain and `beta_(AC)` is `AC` current gain. Also, `A_(V) = beta_(AC) xx` Resistance gain `(=(R_(0))/(R_(i)))` Given, `A_(V) = 50, R_(0) = 200 Omega, R_(i) = 100 Omega` Hence, `50 = beta_(AC) xx (200)/(100)` `:. beta_(AC) = 25` Now, `AC` power gain `= A_(V) xx beta_(AC)` `= 50 xx 25 = 1250` |
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| 15057. |
The susceptibility of magnesium at 300 K is 1.2 x 10–5. At what temperature will the susceptibility increase to 1.8 x 10–5? |
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Answer» Given : x1 = 1.2 x 10-5, T1 = 300 K, x2 = 1.8 x 10-5 To find: Required temperature (T2) Formula: xT = constant Calculation: From formula, x1T1 = x2T2 ∴ T2 = \(\frac{\chi_1T_1}{\chi_2}=\frac{1.2\times10^{-5}\times300}{1.8\times10^{-5}}\) ∴ T2 = 200 K Ans: The required temperature is 200 K. |
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| 15058. |
A voltmeter has a resistance of 100Ω. What will be its reading when it is connected across a cell of e.m.f. 2 V and internal resistance 20Ω? |
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Answer» Given: R = 100Ω, r = 20, E = 2 V To find: Reading of voltmeter (V) Formula: V = E - Ir Calculation: Current through the circuit is given by I = \(\frac{E}{R+r}=\frac{2}{100+20}=\frac{2}{120}\) ∴ I = \(\frac{1}{60} A\) From formula, V = 2 - (\(\frac{1}{60} \times20\)) =2 - 0.3333 ∴ V = 1.667 V Ans: The reading on the voltmeter is 1.667 V. |
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| 15059. |
The figure shows three circuits with identical batteries, inductors and resistance, Rank the circuits according to the currents through the battery just after the switch is closed, greatest first : A. `i_(2) gt i_(3) gt i_(1)`B. `i_(2) gt i_(1) gt i_(3)`C. `i_(1) gt i_(2) gt i_(3)`D. `i_(1) gt i_(3) gt i_(2)` |
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Answer» Correct Answer - A At t = 0, current through inductor is zero At `t = oo`, inductor behave like a conductor |
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| 15060. |
The algebraic sum of all currents meeting at any point in an'electrical circuit is(a) infinite(b) positive (c) zero(d) negative |
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Answer» Correct answer is (c)zero |
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| 15061. |
The algebraic sum of all currents meeting at a point in an electrical circuit is (a) zero (b) infinite (c) positive (d) negative |
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Answer» Correct answer is (a) zero |
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| 15062. |
Inside a supeerconducting ring, six identical resistors each of resistance R are connected as shown in figure. The equivalent resistances. A. between 1 and 3 is zero.B. between 1 and 3 is `R//2`C. between 1 and 2, 2 and 3, 3 and 1 are all equalD. between 1 and 2, 2 and 3 , 3 and 1 are all not equal. |
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Answer» Correct Answer - A::C a., c. Since 1,2, and 3 all at same potential so equivalent resistance between 1 and 2, 2 and 3, 3 and 1 are all zero. |
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| 15063. |
If `a_(1)=a, a_(2)=b, a_(n+2)=(a_(n+1)+a_(n))/2, nge 0` then `lim_(n to oo) a_(n)` is `(a+lamda_(1)b)/(lamda_(2)), (lamda_(1),lamda_(2) epsilonI)` where `lamda_(1)+lamda_(2)` is _______ |
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Answer» Correct Answer - 5 `lim_(nto oo) a_(n)=a+(b-a)/2+(b-a)/8+…=a+2/3(b-a)=(a+2b)/3` |
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| 15064. |
Two cells of emfs `E_1` and `E_2` and of negligible internal resistances are connected with two variable resistors as shown in Fig. A2.4. When the galvanometer shows no deflection, the values of the resistances are P and Q. What is the value of the ratio `E_2/E_1`? ` A. `P/Q`B. `P/(P+Q)`C. `Q/(P+Q)`D. `(P+Q)/P` |
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Answer» Correct Answer - B b. Potential difference `V_P` across P as determined from `E_1` id given by `V_P = (E_1/(P+Q))P.` Potential `V_P` across P as determined from `E_2` is same as `E_2` because no current is drawn, i.e., `V_P = E_2` Therefore, `E_2 = E_1(P/(P+Q)) or E_2//E_1 = (P/(P+Q))` . |
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| 15065. |
Let `alpha and beta` be the roots of `x^2-x-1=0`, with `alpha gt beta`. For all positive integers n, define `a_n=(alpha^n=beta^n)/(alpha-beta),nge 1`. `b_1=1 and b_b=a_(n-1)+a_n+1,n ge 2` Then which of the following options is/are correct ?A. `sum_(n=1)^(oo)(b_n)/(10^n)=(8)/(89)`B. `b_n=alpha^n+beta^n " for all " n ge 1`C. `a_1+a_2+a_3+.......+a_n=a_(n+2)-1" for all " n ge 1`D. `sum_(n=1)^(oo)(a_n)/(10^n)=(10)/(89)` |
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Answer» Correct Answer - B::C::D Given quadratic equation `x^2-x-1=0` having roots `alpha` and `beta,(alpha gt beta)` So, `alpha=(1+sqrt(5))/(2)and beta =(1-sqrt(5))/(2)` and `alpha +beta =1,alpha beta =-1` `because a_n=(a^n-beta^n)/(alpha-beta),n ge1` So, `a_(n+1)=((alpha^n+1)-beta^(n+1))/(alpha-beta)=alpha^n+alpha^n-1beta+alpha^n-2beta+.......alphabeta^(n-1)+beta^(n)` ` =alpha^n-alpha^n-1-alpha^n-3beta......-beta ^n-2+beta^n` `=alpha^n+beta^n-(alpha^n-1+alpha^n-3beta+....+beta^n-2)` `=alpha^n+beta^n-a_(n-1)` `[as a_(n-1)=(alph^n-1 -beta^n-1)/(alpha-beta)=alpha^n-2+alpha^n-3beta+....beta^n-2]` `impliesalpha_(n+1)+alpha_(n-1)=alpha^(n)+beta^(n)=b_(n),AAnge1` So,option (b) is incorrect Now `Sigma_(n=1)^(oo) (b_n)/(10^n)=Sigma_(n=1)^(oo) (a^n+b^n)/(10^n)` ` =Sigma_(n=1)^(oo) ((alpha)/(10))^n+Sigma_(n=1)^(oo) ((beta)/(10))^n [because |(alpha)/(10)|lt 1 and |(beta)/(10)| lt 1 and |(beta)/(10)lt 1|` ` =((alpha)/(10))/(10(alpha)/(10))+((beta)/(10))/(1-(beta)/(10))=(alpha)/(10-alpha)+(beta)/(1-beta)` ` =(10alpha-alpha beta +10beta-alpha beta)/((10-alpha)(10-beta))=(10(alpha+beta)-2alphabeta)/(100-10(alpha+beta)+alpha beta)` `=(10(1)-2(-1))/(100-100(1)-1)` `=(12)/(89)` So, option (a) is not correct. `because alpha^2=alpha +1 and beta^2=beta+1` `rArr alpha^n+1=alpha^n+1+alpha^n and beta^n+2=beta^n+1+beta^n`. `rArr (alpha^n+2+beta^n+2)=(alpha^n+1+beta^n+1)+(alpha^n+beta^n)` `rArr a_(n+1)=a_(n+1)+a_n` Similarly `a_(n+1)=a_n+a_n-1` `a_n=a_n-1+a_(n-2)` `.............` `...................` On adding , we get `a_(n+2)=(a_n+a_n-1+a_n-2+....+a_2+a_1)+a_2` `[because a_2=(alpha^2-beta^2)/(alpha-beta)=alpha+beta=1]` So, ` a_n+2-1=a_1=a_1+a_2+a_3+......+a_n` So, option (c) is aslo correct. And, now `Sigma_(n =1)^(oo) (a_n)/(10^n)=Sigma_(n=1)^(oo) (alpha^n-beta^n)/((alpha-beta) 10^n)` ` =(1)/(alpha-beta)[Sigma _(n=1)^(oo) ((alpha)/(10))^n-Sigma _(n=1)^(oo) ((beta)/(10))^n]` ` =(1)/(alpha-beta) [((alpha)/(10))/(1-(alpha)/(10))-((beta)/(10))/(1-(beta)/(10))],[as |(alpha)/(10)|lt 1 and |(beta)/(10)|lt 1]` `=(1)/(alpha-beta) ((alpha)/(10-alpha)-(beta)/(10-beta))=(1)/(alpha-beta)[(10alpha-alphabeta-10beta+alpha beta)/(100-10(alpha +beta)+alpha beta)]` ` =(10(alpha -beta))/((alpha -beta)[100-10(alpha+beta )+alpha beta])=(10)/(100-10-1)=(10)/(89)` Hence options, (b),(c) and (d) are correct. |
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| 15066. |
A block is hanging at one end of the massless rope passing over a fixed smooth pulley. From other end of the rope a man of same mass climbs with an acceleration "x" relative to the acceleration of rope which is a. The value of "x" is : |
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Answer» Magnitude of acceleration of string equals acceleration of block. |
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| 15067. |
The ammeter has range 1 ampere without shunt. The range can be varied by using different shunt resistances. The graph between shunt resistance and range will have the nature(1) P(2) Q(3) R(4) S |
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Answer» Correct option (2) Q Explanation: As shunt resistance increases, range of ammeter decreases. |
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| 15068. |
`._(29)^(64)` Cu can decay by `beta^(-)` or `beta^(+)` emmision, or electron capture. It is known that `._(29)^(64)` Cu has a half life of `12.8` hrs with 40% probability of `beta^(-)` decay 20% probability of `beta^(+)` decay and 40% probability of electron capture. The mass of `._(29)^(64)` Cu is `63.92977amu whil e `._(30)^(64) Zn is 63.92914 amu` and `._(28)^(64) Ni is 63.92796 amu`. What is the half life for electron capture?A. `5.12` Hrs.B. 32 Hrs.C. `2.56` Hrs.D. 16 Hrs. |
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Answer» Correct Answer - B `._(29)^(64)Cu rightarrow._(30)^(64)Zn+_(-1)^(0)e+vec(V)Q=(m_(Zn)-m_(Cu))C^(2)` `._(29)^(64)Cu rightarrow ._(28)^(64)Ni+._(+1)^(0)e+VQ=(m_(Zn)-m_(Ni)-2me)C^(2)` ` ._(29)^(64)Cu+._(-0)^(0)e rightarrow ._(28)^(64)Ni+VQ=(m_(2n)-m_(Ni))C^(2)` `lambda=lambda_(1)+lambda_(2)+lambda_(3)(lambda_(1)=0.4lambda,lambda_(2)=0.2lambda,lambda_(3)=0.4lambda)` `T_(1//2) e-capture=(T_(1//2))/(0.4)=(12.8)/(0.4)=32` hrs. |
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| 15069. |
`._(29)^(64)` Cu can decay by `beta^(-)` or `beta^(+)` emmision, or electron capture. It is known that `._(29)^(64)` Cu has a half life of `12.8` hrs with 40% probability of `beta^(-)` decay 20% probability of `beta^(+)` decay and 40% probability of electron capture. The mass of `._(29)^(64)` Cu is `63.92977amu whil e `._(30)^(64) Zn is 63.92914 amu` and `._(28)^(64) Ni is 63.92796 amu`. What is the Q value of `beta^(-)` decay?A. `0.587` MeVB. `0.077` MeVC. `1.686` Me VD. `0.666` Me V |
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Answer» Correct Answer - A `Q-(63.92977-63.92914)xx931.5` `=0.00063xx931.5=0.587` MeV |
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| 15070. |
`._(29)^(64)` Cu can decay by `beta^(-)` or `beta^(+)` emmision, or electron capture. It is known that `._(29)^(64)` Cu has a half life of `12.8` hrs with 40% probability of `beta^(-)` decay 20% probability of `beta^(+)` decay and 40% probability of electron capture. The mass of `._(29)^(64)` Cu is `63.92977amu whil e `._(30)^(64) Zn is 63.92914 amu` and `._(28)^(64) Ni is 63.92796 amu`. If initially there was `10^(22)` atoms of`._(29)^(64)` Cu, what is the initial rate at which energy is being produced due to `beta^(+)` decay ?A. `5.8xx10^(4)` WB. `3.2xx10^(3)` WC. `8.4xx10^(2)` WD. `1.6xx10^(4)` W |
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Answer» Correct Answer - B `(dQ)/(dt)=(dN)/(dt)xxQ=(0.666xx10^(6)xx1.6xx10^(-19)xx0.2xxl n 2xx10^(22))/(12.8xx3600)` `=3.2xx10^(3)` W |
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| 15071. |
Two vector `vec(a)=3hat(i)+8hat(j)-2hat(k)` and `vec(b)=6hat(i)+16hat(j)+xhat(k)` are such that the component of `vec(b)`perpendicular to `vec(a)` is zero. Then the value of `x` will be `:-`A. `8`B. `-4`C. `+4`D. `-8` |
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Answer» Correct Answer - B `vec(a)` and `vec(b)` have to be parallel `(3)/(6)=(8)/(16)=(-2)/(x) rArr x=-4` |
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| 15072. |
Vector `vec(A)=hat(i)+hat(j)-2hat(k)` and `vec(B)=3hat(i)+3hat(j)-6hat(k)` are `:`A. ParallelB. AntiparallelC. PerpendicularD. at acute angle with each other |
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Answer» Correct Answer - A Since `vec(B)=3vec(A)`, so both are parallel. |
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| 15073. |
Two forces are acting on a same body `vec(F_(1))=(2hat(i)+3hat(j)-hat(k))` and `vec(F_(2))=(hat(i)+2hat(j)+4hat(k))`, then find component of `vec(F_(1))` along `vec(F_(2))`.A. `((4)/(7sqrt(6)))`B. `(4)/(9sqrt(5))`C. `(4)/(sqrt(21))`D. `(7)/(4sqrt(6))` |
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Answer» Correct Answer - C `vec(F_(1))implies2hat(i)+3hat(j)-hat(k)` `vec(F_(2))implieshat(i)+2hat(j)+4hat(k)` `F_(1)cos thetaimplies (vec(F_(1)).vec(F_(2)))/(|F_(2)|)implies(2+6-4)/(sqrt(21))implies(4)/(sqrt(21))` |
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| 15074. |
If vectors `vec(A)=(hat(i)+2hat(j)+3hat(k))m` and `vec(B)=(hat(i)-hat(j)-hat(k))m` represent two sides of a triangle, then the third side can have length equal to `:`A. `3m`B. `4m`C. `6m`D. Can not be determined |
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Answer» Correct Answer - B Possible length would be `|vec(A)-vec(B)|` or `|vec(A)+vec(B)|` `|vec(A)-vec(B)|=5m`, `|vec(A)+vec(B)|=3m` |
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| 15075. |
Position vector `vec(A)` is `2hat(i)` Position vector `vec(B0` is `3hat(j)` `hat(i),hat(j),hat(k)` are along the shown `x,y,` and `z` is Geometrical representation of `vec(A)` isA. `underset(2 units)rarr`B. `uarr2units`C. `underset(2 units)larr`D. `bar(2 units)` |
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Answer» Correct Answer - A `vec(A)=2hat(i)` `underset(2 units)rarr` |
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| 15076. |
There are two vector `vec(A)=3hat(i)+hat(j)` and `vec(B)=hat(j)+2hat(k)`. For these two vectors- (i) Find the component of `vec(A)` along `vec(B)` and perpendicular to `vec(B)` in vector form. (ii) If `vec(A)` & `vec(B)` are the adjacent sides of parallelogram then find the magnitude of its area. (iii) Find a unit vector which is perpendicular to both `vec(A)` & `vec(B)`. |
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Answer» Correct Answer - (i) `1/5 (hat(j)+2hat(k)), 3hat(i)+4/5hat(j)-2/5hat(k)` (ii) 7 units (iii) `2/7hat(i)-6/7hat(j)+3/7 hat(k)` (i) component of `vec(A)` along `vec(B)=((vec(A).vec(B))/B)hat(B)` `=((vec(A).vec(B))/B)vec(B)/B=[((3hat(i)+hat(j)).(hat(j)+2hat(k)))/sqrt(5)]((hat(j)+2hat(k)))/sqrt(5)=1/5 (hat(j)+2hat(k))` Component of `vec(A) bot vec(B)` `=vec(A)-[(vec(A).vec(B))/vec(B)]hat(B)=3hat(i)+hat(j)-[1/5(hat(j)+2hat(k))]` (ii) Area of the parallelogram `=|vec(A)xxvec(B)|=|(hati,hatj,hatk) ,(3,1,0),(0,1,2)|=|2hati-6hatj+3hatk|` `=sqrt(2^(2)+(-6)^(2)+3^(2))=7` units (iii) Unit vector perpendicular to both `vec(A)` & `vec(B)` `hat(n)=(vec(A)xxvec(B))/(|vec(A)xxvec(B)|)=(2hati-6hatj+3hatk)/7=2/7hat(i)-6/7 hat(j)+3/7 hat(k)` |
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| 15077. |
Position vector `vec(A)` is `2hat(i)` Position vector `vec(B0` is `3hat(j)` `hat(i),hat(j),hat(k)` are along the shown `x,y,` and `z` is Geometrical representation of `vec(B)` is `:`A. `underset(3 units )`B. `uarr3units`C. `underset(3 units)larr`D. `bar(2 units)` |
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Answer» Correct Answer - B `vec(B)=3j uarr3units` |
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| 15078. |
Position vector `vec(A)` is `2hat(i)` Position vector `vec(B0` is `3hat(j)` `hat(i),hat(j),hat(k)` are along the shown `x,y,` and `z` is `-4 vec(A)` can be represented asA. `underset(8 units)rarr`B. `underset(8 units)larr`C. `underset(A)larr`D. `underset(A)rarr` |
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Answer» Correct Answer - B `-4vec(A)=-8iunderset(8 units)larr` |
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| 15079. |
When a compressible wave is sent towards bottom of sea from a stationary ship it is observed that its echo is hear after `2s`. If bulk modulus of elasticity of water is `2xx10^(9)N//m^(2)`, mean temperature of water is `4^(@)` and mean density of water is `1000kg//m^(3)`, then depth of sea will beA. `1014m`B. `1414m`C. `2828m`D. `3000m` |
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Answer» Correct Answer - B `(2d)/(v_(s))=2` `rArrd=v_(s)` `rArr d=sqrt((B)/(rho))=sqrt((2xx10^(9))/(1000))=1414m` |
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| 15080. |
An EM wave is travelling in (i+ j)/√2 direction. Axis of polarization of EM wave is found to be k. Then equation of magnetic field will be |
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Answer» Answer is (2) EM wave is in direction → (i + j)/√2 Electric field is in direction → k E x B → direction of propagation of EM wave |
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| 15081. |
Electric field in space is given by vectorE(t) = E0{(i + j)}/{√2} cos (ωt + Kz). A positively charged particle at (0, 0, π/K) is given velocity v0k at t = 0. Direction of force acting on particle is(1) f = 0(2) Antiparallel to {(i + j)}/{√2}(3) Parallel to {(i + j)}/{√2}(4) k |
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Answer» Answer is (2) Force due to electric field is in direction - {i + j}/{√2} because at t = 0, E = - {i + j}/{√2}E0 Force due to magnetic field is in direction q(vectorv x vectorB) and vectorv || k ∴ it is parallel to vectorE ∴ net force is antiparallel to {i + j}/{√2} |
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| 15082. |
A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by vector B = 5 x 10-8 jT. The corresponding electric field vector E is (speed of light c = 3 × 108 ms–1)(1) 1.66 × 10–16 iV/m(2) 15 iV/m(3) -1.66 x 10-16 iV/m (4) -15 iV/m |
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Answer» Answer is (2) 15i V/m E/B = c E = B × c = 15 N/c Answer is (2) 15 iV/m E = vector (B x V) = (5 x 10-8 j) x (3 x 108 k) = 15 iV/m |
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| 15083. |
A composite wire is made by joining two uniform wires. If `l_1 = l_2= l` and ` mu_1 = mu_2 /9 = mu`. Tension in the strings is T , `mu` is mass per unit length. Then lowest frequency such that the junction is an antinode. A. `1/(4l) sqrt(l/(mu))`B. `3/(4l)sqrt(l/(mu))`C. `5/(4l) sqrt(l/(mu))`D. `7/(4l)sqrt(l/(mu))` |
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Answer» Correct Answer - A For antinode `f_(1)=((2n_(1)-1))/(4l)sqrt(l/(mu))` `f_(2)=((2n_(2)-1))/(4l)sqrt(l/(9mu)) =((2n_(2)-1))/(12l)sqrt(l/(mu))` `f=1/(4l)sqrt(l/(mu)) ((2n_(1)-1))/(4l)=(2n_(2)-1)/(12l)` `6n_(1)-3 =2n_(2) -1` `6n_(1)-2n_(2) =2 " " n_(1)=1, n_(2)=2` `3n_(1)-n_(2)=1`, |
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| 15084. |
If the lines `L_1:2x-3y-6=0,L_2: x+y-4=0`and `L_3: x+2`and `tanA ,tanB`and `tanC`are the roots of the equation `a x^3+b x^2+c x-15=0`then`a=3`(b) `b=-15``c=28`(d) `a+b+c=15`A. `sum tan A = 15/2`B. `sum tan A tan B = 14`C. `tanA tan B tan C = 15/2`D. equation whose roots are `tan A, tan B, tan C` is `2x^(3)- 15x^(2) +28x -15 =0` |
| Answer» Correct Answer - A | |
| 15085. |
A particle of charge per unit mass `alpha` is released from origin with a velocity `vecv=v_(0)hati` uniform magnetic field `vecB=-B_(0)hatk`. If the particle passes through `(0,y,0)`, then `y` is equal toA. `-(2v_(0))/(B_(0)alpha)`B. `(v_(0))/(B_(0)alpha)`C. `(2v_(0))/(B_(0)alpha)`D. `-(v_(0))/(B_(0)alpha)` |
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Answer» Correct Answer - C Directin of force along `+y` direction so partical passes throught the `+y` axis at a `y = 2r = (2mv)/(rhoB) = (2 v_(0))/(B_(0) alpha)` |
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| 15086. |
The capacitance of a cylindrical capacitor can be increased by: A. decreasing both the radius of the inner cylinder and the length B. increasing both the radius of the inner cylinder and the length C. increasing the radius of the outer cylindrical shell and decreasing the length D. decreasing the radius of the inner cylinder and increasing the radius of the outer cylindrical shell E. only by decreasing the length |
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Answer» B. increasing both the radius of the inner cylinder and the length |
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| 15087. |
A 2-µF and a 1-µF capacitor are connected in series and a potential difference is applied across the combination. The 2-µF capacitor has: A. twice the charge of the 1-µF capacitor B. half the charge of the 1-µF capacitor C. twice the potential difference of the 1-µF capacitor D. half the potential difference of the 1-µF capacitor E. none of the above |
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Answer» D. half the potential difference of the 1-µF capacitor |
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| 15088. |
A 2-µF and a 1-µF capacitor are connected in parallel and a potential difference is applied across the combination. The 2-µF capacitor has: A. twice the charge of the 1-µF capacitor B. half the charge of the 1-µF capacitor C. twice the potential difference of the 1-µF capacitor D. half the potential difference of the 1-µF capacitor E. none of the above |
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Answer» A. twice the charge of the 1-µF capacitor |
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| 15089. |
A flase balance has equal arms. An object weights `W_1` when placed in one pan and `W_2` when placed in the other pan. The true weight `W` of the object is.A. `sqrt(xy)`B. `(x+y)/(2)`C. `(x^(2)+y^(2))/(2)`D. `(sqrt(x^(2)+y^(2)))/(2)` |
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Answer» Correct Answer - B Let the weights of the pans be `W_(1)` and `W_(2)` and true weight be `W`. Then `(W+W_(1))l=(x+W_(2))l`…..(`1`) and `(W+W_(2))l=(y+W_(1))l`…….(`2`) `implies W=(x+y)/(2)` |
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| 15090. |
iii) A circular loop of area \( \pi cm ^{2} \) and carrying current \( 10 \mu A \) is placed in magnetic field \( B \) with its plane perpendicular to \( B \). If \( B=\pi mT \), the magnitude of torque exerted on this loop isa) zerob) \( 10^{-11} Nm \)c) \( 10^{-7} Nm \)d) \( 10^{-1} Nm \) |
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Answer» Answer is (a) 0 Given area A = π cm2 i = 10 μA B = π mT θ = 0 We know that magnitude of torque τ = N i A B sin θ τ = 1 x 10 x 10-6 x π x 10-4 x π x sin 0° τ = 0 (∵ sin θ = 0) |
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| 15091. |
A particle of charge `-q` and mass m moves in a circle of radius r around an infinitely long line charge of linear charge density `+lamda`. Then, time period will be where , `k=1/4piepsilon_0)`A. `T=2pi rsqrt((m)/(2k lambdaq))`B. `T^(2)=(4pi^(2)m)/(2k lambda q)r^(3)`C. `T=(1)/(2 pi r) sqrt((2k lambdaq)/(m))`D. `T=(1)/(2pi r) sqrt((m)/(2k lambdaq))` |
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Answer» Correct Answer - A We have centripetal force equation `q((2k gamma)/(r))=(mv^(2))/(r)` so `v=sqrt((2kq gamma)/(m))` Now `T=(2pi r)/(v)=2pirsqrt((m)/(2 k gamma q))` where, `k=(1)/(4pi epsilon_(0))` |
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| 15092. |
A thin prism of glass is placed in air and water respectively. If `n_(0) = (3)/(2)` and `n_(w) = (4)/(3)`, then the ratio of deviation produced by the prism for a small angle of incidence when placed in air and water separately is:A. `9 : 8`B. `4 :3`C. `3 :4`D. `4 :1` |
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Answer» Correct Answer - D `delta_(a)=((3)/(2)-1)xx A=(A)/(2)` `delta_(w)=((3//2)/(4//3)-1)A=(A)/(8)` `(delta_(air))/(delta_(water))=(4)/(1)`. |
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| 15093. |
A graph of the x-component of the electric field as a function of x in a region of space is shown in figure. The `y- and z-components ` of the electric field are zero in this region. If the electric potential is 10 V at the origin, then the potential at x = 2.0 m is A. 10VB. 40VC. `-10V`D. 30V |
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Answer» Correct Answer - D `v_(B)-v_(A)=-intE_(x)dx=-` [Area under `E_(x)-x` curve] `v_(B)-10=-(1)/(2).2.(-20)=20` `v_(B)=30V`. |
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| 15094. |
A graph of the x-component of the electric field as a function of x in a region of space is shown in figure. The `y- and z-components ` of the electric field are zero in this region. If the electric potential is 10 V at the origin, then the potential at x = 2.0 m is A. `10 V`B. `40 V`C. `-10 V`D. ` 30 V` |
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Answer» Correct Answer - D `V_(B)-V_(A)=-intE_(X)dx=-["area under "E_(x)-x," curve"]` `V_(B)-10=-(1)/(2)xx2xx(-20)=20` `V_(B)=30V` |
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| 15095. |
A light ray incident along vector `2hati+4hatj+sqrt(5)hatk` strikes in the `x-z` from medium of refractive index `sqrt(3)` and enters into medium II of refractive index is `mu_(2)`. The value of `mu_(2)` for which the value of angle of refraction becomes `90^(@)` isA. `(4sqrt(3))/5`B. `(3sqrt(3))/5`C. `(2sqrt(3))/5`D. `(sqrt(3))/5` |
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Answer» Correct Answer - B `sintheta_(C)=mu_(2)//mu_(1)` `1-cos^(2)theta_(C)=((mu_(2))/(mu_(1)))^(2)` `1-(hatn.hatp)^(2)=((mu_(2))/(mu_(1)))^(2)` `1-[4/(sqrt(25))]^(2)=((mu_(2))/(mu_(1)))^(2)implies(mu_(2))/(mu_(1))=3/5` `implies mu_(2)=(3sqrt(3))/5` |
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| 15096. |
Two infinite parallel,non- conducting sheets carry equal positive charge density `sigma` ,One is placed in the `yz` plane other at distance `x=a` Take potential `V=0` at `x=0` thenA. For `0 lt x lt a`, potential `V = 0`.B. For `x gt a`, potential `V = -(sigma)/(in_(0))(x-a)`C. For `x gt a`, potential `V = (sigma)/(in_(0))(x-a)`D. For `x lt 0` potential `V = (sigma)/(in_(0))x` |
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Answer» `0ltxlta:V=[-int_(0)^(x)E_(x)dx]+V_((0))" "=0`(as `E_(x)=0`) `xgta,V=-int_(a)^(x)E_(x)dx+V_((a))" "=[-int_(a)^(x)(sigma)/(in_(0))dx]+V_((a))=-(sigma)/(in_(0))(x-a)` `xlt0,V=-int_(0)^(x)E_(x)dx+V_((0))" "=-(-(sigma)/(in_(0)).x)+V_((0))=(sigma)/(in_(0)).x` |
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| 15097. |
Given Rydberg costant` R = 10^(5)cm^(-1)`. Supposing if electron jumps from M shell to K shell of H - atom, the frequency of the radiation emitted in cycle/s would be :-A. `(8)/(9)xx10^(5)`B. `(8)/(3)xx10^(15)`C. `(8)/(3)xx10^(11)`D. `(8)/(9)xx10^(15)` |
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Answer» Correct Answer - B `u=RCZ^(2)((1)/(1^(2))-(1)/(3^(2)))` `=10^(7)xx3xx10^(8)xx(8)/(9)=(8)/(3)xx10^(15)` |
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| 15098. |
Two identical dielectric slabs, A and B, are placed symmetrically between the plates X and Y of charged parallel-plate capacitor. The electric intensity has magnitudes `E_(1)//E_(2)` and `E_(3)` at the points `1, 2` and `3`A. `E_(1) gt E_(2) gt E_(3)`B. `E_(1) = E_(3) lt E_(2)`C. `E_(1) = E_(3) gt E_(2)`D. `E_(1) = E_(2) = E_(3)` |
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Answer» Correct Answer - D |
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| 15099. |
Why does a dry cell become dead even if it has not been used for a long time? |
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Answer» NH4 Cl is acidic in nature. It corrodes zinc container. |
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| 15100. |
Prove that ABCD given in the figure is cyclicDraw figure and mark PQ If ∠BAP = x then what is ∠BQP? Find ∠PQD Find ∠PDC? Why? What is ∠A + ∠C? |
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Answer» i. Quadrilateral ABPQ is cyclic If ∠A = x, then ∠BQP = 180 – x If ∠B = y, then ∠APQ = 180 – y Quadrilateral PQCD is cyclic, SO ∠QCD = 180 – x (∠DPQ = x ) ∠PDC = 180 – x (∠PQC = y) ∴ ABCD is a cyclic quadrilateral. |
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