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InterviewSolution
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.
| 1. |
If `veca+vecb+vecc=vec0, |veca| = 3, |vecb| = 5, |vecc| = 7`, then angle between `veca` and `vecb` isA. ` ( pi ) /( 2 ) `B. ` ( pi ) /( 3 ) `C. ` ( pi ) /( 4 ) `D. ` ( pi ) /( 6 ) ` |
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Answer» Correct Answer - B Given , a + b + c = 0 and ` |a| = 5, |b| = 3, |c| = 7 ` ` rArr a + b = - c ` On squaring both sides, we get ` (a + b ) ^ 2 = ( - c ) ^ 2 ` `rArr |a + b| ^ 2 = |c| ^ 2 ` ` rArr |a| ^ 2 + |b| ^ 2 + 2a * b = |c| ^ 2 ` ` ( because theta ` be the angle between a and b ) ` rArr ( 5 ) ^ 2 + ( 3 ) ^ 2 + 2| a||b| cos theta = ( 7 ) ^ 2 ` ` rArr 25 + 9 + 2* 5 * 3 * cos theta = 49` ` rArr 30 cos theta = 15 ` ` rArr cos theta = ( 1 ) / ( 2 ) = cos 60 ^@ ` `rArr theta = ( pi) / ( 3 ) ` |
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| 2. |
If the position vectors of the vertices A, B and C are 6i, 6j and k respectively w.r.t origin O, then the volume of the tetrahedron OABC isA. ` 6 `B. ` 3 `C. ` ( 1 )/( 6 ) `D. ` (1 )/( 3 ) ` |
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Answer» Correct Answer - A Given that, the position vectors of the vertices A, B and C are `OA = 6i= 6i+0j+0k ` OB =6j=0i+6j+0k OC = k=0i+0j +k Now, volume of the tetrahedron ` = ( 1 )/ ( 6 ) ` [OA OB OC] = ` ( 1 ) / ( 6 ) |{:( 6,,0,,0),(0,,6,,0), ( 0,,0,,1):}| ` = ` ( 1 ) / ( 6 ) ( 6 xx 6 xx 1 ) = 6 ` |
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| 3. |
`i(jxx k ) + j ( k xx i) + k (j xx i) ` is equal toA. ` 3 `B. ` 2 `C. `1`D. `0` |
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Answer» Correct Answer - C Given `i(j xx k ) + j ( k xx i) + k(j xxi) ` ` i*(i) + j*(j) + k(-k) ` ` = (i*i) + (j*j) - (k*k) ` ` = 1 + 1 + 1 - 1 = 1 ` |
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| 4. |
The value of `int_0^oox/((1+x)(x^2+1))dx` isA. ` 2pi `B. ` ( pi ) /( 4 ) `C. ` ( pi ) /( 16 ) `D. ` ( pi ) /( 32 ) ` |
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Answer» Correct Answer - B Let ` I = int _ 0 ^( oo) ( x dx ) /( (1 + x ) ( x^ 2 + 1 ) ) ` by partial fraction, ` ( x )/( ( 1 + x ) ( x^ 2 + 1 ) ) = ( A) / ( ( 1 + x ) ) + (B x + C ) /( ( x ^ 2 + 1 )) ` ` rArr x = A ( x^ 2 + 1 ) + ( 1 + x ) (Bx +C ) ` ` rArr x= A ( x^ 2 + 1 ) + ( B x + Bx ^ 2 + C + Cx ) ` ` rArr x = ( A +B ) x^ 2 + ( B + C ) x + ( A+C ) ` On comparing both sides , we get ` A + B = 0 , B +C = 1, A +C= 0 ` ... (i) On adding all these equations , we get ` A + B+C = ( 1 ) / ( 2 ) " " `... (ii) ` therefore A = (1 ) / ( 2 ) - 1 = - ( 1 ) / ( 2 ) , C= ( 1 )/ ( 2 ) ` and ` B = ( 1 ) / (2 ) ` ` therefore I = int _ 0 ^( oo) { ( - 1 ) /( 2 ( 1 + x ) ) + ( 1 ) / ( 2) ( ( x + 1 ))/ ( ( x^ 2 + 1 ))} dx ` ` = - ( 1 ) / ( 2 ) int _ 0 ^( oo) ( dx ) / ( 1 + x ) + (1 ) / ( 2 ) int _ 0 ^( oo) ( x ) /( x ^ 2 + 1 ) dx + ( 1 ) / (2 ) int _ 0^(oo) ( dx ) / ( 1 + x ^ 2 ) ` ` =- (1 ) / ( 2 ) [log ( 1 + x ) ]_0 ^(oo) + (1 )/ ( 4 ) [log ( x ^ 2 + 1 ) ] _0 ^(oo) + ( 1 )/ (2 ) xx ( pi ) / (2 ) ` ` = - ( 1 ) / ( 2 ) lim _ ( xto oo ) log ( 1 + x ) + (1 )/ ( 4 )` ` lim_ ( x to oo) log ( 1 + x ^2 ) + ( pi )/(4 ) ` ` = lim _ ( xto oo) log [ ( ( 1 + x^ 2 ) ^ (1//4) ) / ( ( 1 + x ) ^ ( 1// 2 ) ) ] + ( pi ) /( 4 ) ` ` = lim_ ( x to oo ) log [ (sqrt x (( 1 )/ ( x^ 2) + 1 ) ^ (1// 4) ) / ( sqrt x ( ( 1 )/ ( x ) + 1 )^ (1// 2 ) ) ] + (pi )/ ( 4 ) ` ` =log ""( ( 0 + 1 ) ^ (1//4))/( ( 0 + 1 )^(1//2 ) ) + ( pi) / ( 4 ) ` ` = log ( 1 ) + ( pi ) / (4 ) = 0 + (pi ) /(4 ) = ( pi ) / ( 4 ) ` |
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| 5. |
The value of `int (dx)/(x(x^(n)+1))`A. ` ( 1 ) /( n ) log (( x^ n ) /( x ^n + 1 )) + C `B. `log (( x ^n + 1 ) /( x^n )) + C `C. ` ( 1 )/( n ) log (( x^n + 1 ) /( x^n )) + C `D. `log (( x^n ) /( x^n + 1 )) + C ` |
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Answer» Correct Answer - A Let I ` = int ( dx ) / ( x ( x^n + 1 )) ` (let ` t = x^n + 1, dt = nx ^ ( n - 1 ) dx` ) ` = int ( dx ) /( nx^n * t ) " " ( ( dt ) /( nx ^n ) = ( dx ) / ( x )) ` ` = ( 1 )/ ( n ) int (dt ) /( t ( t - 1 )) ` ` = ( 1 )/( n ) int { ( 1 )/( t - 1 ) - ( 1 )/ (t )} dt ` ` = ( 1 )/ (n ) { log ( t- 1 ) - log t } + C ` `= ( 1 )/ ( n ) log (t - 1 ) /( t ) +C ` `( 1 )/( n ) log ( (x^n)/(x^n + 1 ) + C ` |
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| 6. |
Find a vector of magnitude 9, which is perpendicular to both vectors `4 hat i- hat j+3 hat ka n d-2 hat i+ hat j-2 hat k`.A. `3i + 6j - 6 k `B. ` 3i - 6j + 6k `C. `-3i + 6j + 6k `D. None of the above |
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Answer» Correct Answer - C Let ` a = 4i - j + 3k, b = - 2i + j - 2 k ` and ` c =x i + yj + zk ` Given ` a * c = 0 ` i.e., ` 4 x -y + 3z= 0 " " `… (i) and ` b * c = 0 ` i.e., ` - 2x + y -2z = 0 " " `...(ii ) Also, `|c | = 9 ` i.e., ` x ^ 2 + y ^ 2 + z ^ 2 = 81 " " `... (iii) Now, from Eqs. (i) and (ii), we get ` 2x + z = 0 rArr z = - 2x ` On putting this value in Eq. (iii), we get ` x ^ 2 + y ^ 2 + 4 x^ 2 = 81 ` ` rArr 5x^ 2 + y ^ 2 = 81 " " `... (iv) On multiplying Eq. (i) by 2 and Eq. (ii) by 3 and then adding , we get ` {:( 8x - 2y + 6 z= 0 ), (ul(-6x + 3y - 6z = 0 )), ( 2x + y = 0 ) :} ` ` rArr y = - 2x ` On putting this value in Eq. (iv) , we get ` 5x ^ 2 + 4x ^ 2 = 8 1 ` ` rArr 9 x ^ 2 = 81 ` ` rArr x ^ 2 = 9 ` ` rArr x = pm 3 ` ` therefore y = pm 6 and z = pm 6 ` ` therefore ` Required vector, ` c = x i + yj + zk ` ` = pm 3i pm 6j pm 6 k ` ` = 3i - 6 j - 6k ` ` = -3i + 6j + 6k` |
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| 7. |
`int_0^pilog(1+cosx)dx` .A. ` - ( pi ) /( 2 ) log 2 `B. ` pi log "" ( 1 ) /( 2 ) `C. ` pi log 2 `D. ` ( pi ) /( 2 ) log 2 ` |
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Answer» Correct Answer - B Let I ` = int _ 0 ^ ( pi) log ( 1 + cos x ) dx " " `… (i) I = ` int _ 0 ^( pi ) log { 1 + cos (pi- x )} dx ` ` = int_ 0 ^(pi ) log ( 1 - cos x ) dx " " `… (ii) On adding Eqs. (i) and (ii), we get ` 2I = int _ 0 ^( pi ) {log ( 1 + cos x ) + log ( 1 - cos x ) } dx ` ` I = ( 1 ) /( 2 ) int _ 0 ^( pi) log ( 1 - cos ^( 2 ) x ) dx ` ` = ( 1 ) /( 2 ) int _0 ^(pi) log sin ^2 x dx ` ` = int _ 0^(pi ) log sin x dx ` ` = 2 int _ 0 ^(pi // 2 ) log sin x dx ` ` { because int _0 ^ ( 2a ) f ( x ) dx = 2 int_ 0 ^ ( a ) f (x ) dx, if f ( 2a - x ) =f ( x ) } ` ` = 2{ - ( pi ) / ( 2 ) log 2} ` ` ( because int _ 0 ^( pi//2 ) log sin x dx = - ( pi ) /( 2 ) log 2 ) ` ` = pi log"" ( 1 )/ ( 2 ) ` |
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| 8. |
The value of `int (dx)/(x(x^(n)+1))`A. ` ( 1 ) /( n ) log (( x^ n ) /( x ^n + 1 )) + C `B. `log (( x ^n + 1 ) /( x^n )) + C `C. ` ( 1 )/( n ) log (( x^n + 1 ) /( x^n )) + C `D. `log (( x^n ) /( x^n + 1 )) + C ` |
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Answer» Correct Answer - A Let I ` = int ( dx ) / ( x ( x^n + 1 )) ` (let ` t = x^n + 1, dt = nx ^ ( n - 1 ) dx` ) ` = int ( dx ) /( nx^n * t ) " " ( ( dt ) /( nx ^n ) = ( dx ) / ( x )) ` ` = ( 1 )/ ( n ) int (dt ) /( t ( t - 1 )) ` ` = ( 1 )/( n ) int { ( 1 )/( t - 1 ) - ( 1 )/ (t )} dt ` ` = ( 1 )/ (n ) { log ( t- 1 ) - log t } + C ` `= ( 1 )/ ( n ) log (t - 1 ) /( t ) +C ` `( 1 )/( n ) log ( (x^n)/(x^n + 1 ) + C ` |
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| 9. |
The value of ` int ( ( x ^ 2 + 1 )) /( x ^ 4 + x ^ 2 + 1 ) dx ` isA. ` (1 ) /( sqrt 3 ) tan ^( - 1 ) { ( x - 1// x ) /( sqrt 3)} + C `B. ` ( 1 ) /( 2sqrt 3 ) log { ( ( x- 1 // x) - sqrt 3) /( (x - 1 // x ) + sqrt 3 } + C `C. ` tan ^( - 1 ) (( x + 1// x ) /(sqrt 3) ) + C `D. `tan^( - 1 ) (( x - 1//x )/ (sqrt 3 )) + C ` |
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Answer» Correct Answer - A Let ` I= int ( x^ 2 + 1 ) /( x ^4 + x^ 2 + 1 ) dx ` ` = int (( 1 + ( 1) / ( x^ 2 )))/ ( ( x ^ 2 + 1 + ( 1 ) / ( x ^ 2 )) ) dx ` ` = int (( 1 + ( 1 )/ ( x ^ 2 ))) /( ( x - ( 1 )/( x) ) ^ 2 + (sqrt 3 )^ 2 ) dx ` ` = int (dt )/( ( sqrt 3 )^ 2 + t^ 2 ) ` [ let ` t = x - ( 1 ) / ( x ) rArr dt = ( 1 + ( 1 )/ ( x ^ 2 ) ) dx] ` ` = ( 1 )/(sqrt 3 ) tan ^( - 1 ) (( t )/ (sqrt 3 ) ) + C ` ` = ( 1 )/ ( sqrt 3 ) tan^( - 1 ) { ( 1 )/ (sqrt3 ) ( x - ( 1 )/ (x )) } + C ` |
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| 10. |
`int sin(2x)/(sin^4x+cos^4x) dx`A. ` tan ^( - 1 ) (cot ^ 2 x ) + C `B. ` - tan ^( - 1 ) ( cos 2x ) + C `C. ` tan ^( - 1 ) ( sin 2x ) + C `D. ` tan ^( -1 ) ( tan ^ 2 x) +C ` |
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Answer» Correct Answer - B Let I ` = int ( sin 2x ) /( sin ^ 4 x + cos ^ 4 x ) dx ` ` I = int ( sin 2x )/( ( sin ^ 2 x + cos ^ 2 x ) ^ 2 - 2 sin ^ 2 x * cos ^ 2 x ) dx ` I ` = int ( sin 2x )/ ( 1 - (1 )/ (2 ) ( sin 2x )^ 2 ) dx ` ` = 2 int ( sin 2 x ) /( 1 + ( 1 - sin ^ 2 2x ) ) dx ` (let ` t = cos 2x rArr dt = - 2 sin 2x dx ` ` = int ( -dt )/( 1 + t ^ 2 ) ` ` =- tan ^( - 1) t + C ` ` =-tan ^( - 1 ) ( cos 2x) + C ` |
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| 11. |
The value of `int(1)/(3 sin x -cos x +3)dx` is equal toA. ` log (( tan "" ( x ) /( 2 ) + 1 ) /( 2 tan "" ( x ) /( 2) + 1 ) ) +C `B. ` ( 1 ) /( 2 ) log ( ( 2tan "" ( x ) /( 2 ) + 1 ) / ( tan "" ( x ) /(2 ) +1 ) ) + C `C. `log ( ( 2tan "" ( x ) /( 2 ) + 1 ) /( tan "" ( x ) /( 2 ) + 1 ) ) + C `D. ` 2log (( 2 tan "" ( x ) /( 2 ) + 1 ) /( tan "" ( x ) /( 2) + 1 ) ) +C ` |
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Answer» Correct Answer - C Let I ` = int (dx )/ ( 3 sin x - cos x + 3 ) ` ` {{:( because, sin x = (2 tan "" ( x ) / ( 2 )) /( 1 + tan ^ 2 "" ( x )/ ( 2 )) ) , ( and , cos x = ( 1 - tan ^ 2 "" ( x )/(2 ))/(1 +tan ^( 2) "" ( x )/ ( 2 ))):}} ` ` I = int ( dx ) / ( 3 { ( 2 tan "" (x ) / ( 2 ) ) / ( 1 + tan ^ 2 "" ( x )/ ( 2 ) )} - { ( 1 - tan ^ 2 "" ( x ) / ( 2 ))/ ( 1 + tan ^ 2"" ( x )/ ( 2 ))} + 3 ) ` ` = int (( 1 + tan ^ 2 "" ( x )/ ( 2 )) dx )/ ( 6 tan "" ( x )/ ( 2 ) - 1 + tan ^ 2 "" ( x ) / ( 2 ) + 3 + 3tan ^( 2 ) "" ( x ) / ( 2 ) ) ` ` = int (sec^ 2 "" ( x ) / ( 2 ) ) / ( 4 tan ^ 2 "" (x ) / ( 2 ) + 6 tan"" (x )/ (2 ) + 2) dx ` ( let ` t = tan "" ( x )/ ( 2 ), dt= ( 1 )/ (2) sec^ 2 "" ( x )/ (2 ) dx` ) ` = int ( dt ) / ( 2 t^ 2 + 3t + 1 ) ` ` = int ( dt )/ ( ( t + 1 ) ( 2t+ 1 ) ) ` ` = int { ( - 1 )/ ( ( t + 1 ) ) + ( 2 ) /( ( 2t + 1 ))} dt ` ( by partical fraction ) ` = -log ( t + 1 ) + ( 2) / ( 2 ) log ( 2 + 1 ) + C ` ` = log"" (( 2t + 1 ))/( (t + 1 )) + C ` ` = log (( 2 tan"" ( x )/ ( 2 ) + 1 )/ ( tan "" ( x ) / ( 2 ) + 1 ) ) + C ` |
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| 12. |
The value of ` int _ 3 ^ (4 ) sqrt((4- x ) (x - 3 )) dx ` isA. ` ( pi ) /( 16 ) `B. ` ( pi ) /( 8 ) `C. ` ( pi ) /( 4 ) `D. ` ( pi ) /( 2 ) ` |
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Answer» Correct Answer - B Let `I = int_ 3^ 4 sqrt ((4 - x ) ( x - 3)) dx ` ` = int _ 3 ^ 4 sqrt ( - x ^ 2 + 7x -12 ) dx ` ` = int_ 3 ^ 4 sqrt ( - ( x ^ 2 - 7x + (49)/ ( 4 ) - ( 49)/( 4 )) - 12 ) dx ` = ` int_ 3 ^ 4 sqrt ( - ( x - ( 7 ) /( 2 ) ) ^ 2 + ( 49 ) /( 4) - 12 ) dx ` ` = int _ 3 ^ 4 sqrt ( ( 1 ) / ( 4 ) - ( x - ( 7 ) / ( 2)) ^ 2 ) dx ` Let ` t = x - ( 7 ) /( 2 ) rArr dt = dx ` ` therefore ` Upper limit = ` ( 1) / ( 2 ) ` and lower limit = - ` ( 1 ) / (2 ) ` ` = int _ ( - 1//2 ) ^ ( 1//2 ) sqrt (( ( 1 )/( 2 ) ) ^ 2 - t ^ 2 ) dt ` ` = 2 int _ 0 ^ ( 1//2 ) sqrt(( ( 1 )/( 2))^ 2 - t ^ 2 ) dt ` `= 2 [ ( t ) /(2) sqrt (( 1 ) /( 4) - t ^ 2 ) + ( 1 ) /( 8 ) sin ^( - 1 ) 2t] _ ( 0) ^( 1//2 ) ` ` = 2 [0 + ( 1 ) /(8) xx ( pi ) / ( 2 )] ` ` = ( pi )/( 8 ) ` |
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| 13. |
The oxidation number of C atom in ` CH _ 2 Cl _ 2 and CCl_ 4 ` are respectivelyA. ` - 2 and 4 `B. ` 0 and - 4 `C. ` 0 and 4 `D. ` 2 and 4 ` |
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Answer» Correct Answer - C Let the oxidation number of C is x ` CH _ 2 Cl _ 2 ` ` x + 2 ( + 1 ) + 2 ( -1 ) = 0 ` ` x + 2 - 2 = 0 ` ` x = 0 ` ` C Cl _ 4 ` ` x + 4 ( - 1 ) = 0` ` x - 4 = 0` ` x = 4 ` |
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| 14. |
` C _ 6 H _ 6 + CH _ 3 Cl overset ( "Anhy. " AlCl _ 3 ) to C _ 6 H _ 5 CH _ 3 + HCl ` the name of the above reaction isA. GattermannB. Reimer - TiemannC. Friedel - craftD. Cannizzaro |
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Answer» Correct Answer - C ` C_ 6 H _ 6 + CH _ 3Cl overset ("Anhy. " AlCl _ 3 ) to C _ 6 H _ 5CH _3 + HCl ` This reaction is known as Friedel - Craft reaction. |
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| 15. |
In AC circuits choke is preferred to resistors becauseA. choke coil is cheapB. voltage increasesC. energy is not wastedD. current increases |
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Answer» Correct Answer - C In AC circuits choke coil is used to control current is without loss of energy. |
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| 16. |
AC measuring instruments measuresA. peak valueB. Average valueC. any valueD. RMS value |
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Answer» Correct Answer - D AC measuring instrument ( AC ammeter and voltemeter ) always measures rms value. |
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| 17. |
Which of the following compound of xenon does not exists ?A. ` Xe F _ 6 `B. `Xe F _ 4 `C. `XeF _ 5 `D. `XeF _ 2 ` |
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Answer» Correct Answer - C Xenon forms following fluorides ` Xe F _ 2 , X e F _ 4 and Xe F _ 6 ` It does not exists as ` Xe F _ 5 `. |
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| 18. |
The strongest acid isA. acetic acidB. trichloracetic acidC. dichloracetic acidD. monochloroacetic acid |
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Answer» Correct Answer - B Since, chloro group has - I effect. Due to which greater the number of electron withdrawing atoms (halogens ) , stronger will be the acid. Thus, order of acidity is ` Cl - overset ( Cl) overset (|) underset (Cl) underset (| )C - COOH gt H - overset (Cl) overset (| )underset(Cl) underset(|)C - COOH gt ` trichloroacetic acid `" " ` dichloroacetic acid `H - overset (Cl) overset(|)underset(H) underset(|)C - COOH gt H - overset(H)overset(|) underset(H)underset(|)C -COOH ` chloroacetic acid `" " ` acetic acid |
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| 19. |
The number of alkanyl groups possible from ` C_ 4H _ 7 ` areA. 7B. 5C. 3D. 8 |
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Answer» Correct Answer - B Following five alkenyl groups are possible from ` -C _4 H _ 7 ` (i) ` CH _ 3 - CH _ 2 -CH =CH - ` (ii) `CH _3 -CH = CH-CH _ 2 - ` (iii ) ` CH _ 2 =CH -CH _ 2 -CH _ 2 - ` (iv) ` CH _ 3- underset (CH _ 3) underset(|) C -CH - ` (v) ` CH _3 - underset ( CH _ 2- ) underset ( |) C =CH ` |
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| 20. |
The solution of ` BiCl _ 3 ` in dil. HCl when diluted with water , white precipitate is formed which isA. Bismith oxychlorideB. Bismith oxideC. Bismith hydroxideD. None of the above |
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Answer» Correct Answer - A ` BiCl _ 3 + H _ 2 O hArr underset ("(white precipitate)") underset("Bismuth oxy chloride")(BiOCl downarrow + 2HCl ) ` |
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| 21. |
If ` 2a + b + 3c = 0` , then the line ` a x + by + c = 0 ` passes through the fixed point that isA. ` (( 2 )/( 3), ( 1 ) /( 3)) `B. ` (0, 1 ) `C. ` ( (2 )/( 3), 0 ) `D. None of these |
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Answer» Correct Answer - A Given, ` 2a + b + 3c = 0 " " ` … (i) and line, ` ax + by +c= 0 ` ` rArr 3ax + 3by + 3c = 0 " " `… (ii) On subtracting Eq. (i) from Eq. (ii), we get ` ( 3x -2 ) a + (3y - 1 ) b = 0 * a + 0* b ` On comparing both sides , we get ` 3x - 2 = 0 rArr x = (2 )/( 3 ) ` ` and 3y - 1 = 0 rArr y = ( 1 ) / (3 ) ` ` therefore ` Line ` ax + by + c = 0 ` passes through the fixed point ` (( 2 ) /( 3 ) , ( 1 ) /( 3 ) ) ` |
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| 22. |
A body is projected up along a rough inclined plane of inclination ` 45 ^@ ` . The coefficient of friction is 0.5. Then the retardation of the block isA. ` ( g ) /( 2 sqrt 2 ) `B. ` ( g ) /(2 ) `C. ` ( 3g )/( 2sqrt 2 ) `D. ` ( g ) /( sqrt 2 ) ` |
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Answer» Correct Answer - C Retarding ` = g ( sin theta = mu cos theta ) ` ` = g ( (1 )/( sqrt 2 ) + 0.5 ( 1 )/( sqrt 5 ) ) ` ` = ( g )/( sqrt 2 ) ( 1 + 0.5 ) ` ` = ( 1.5 g )/( sqrt2 ) = ( 3g )/( 2 sqrt 2 ) ` |
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| 23. |
Two coils of self-inductance `L_(1)` and `L_(2)` are placed closed to each other so that total flux in one coil is completely linked with other. If `M` is mutual inductance between them, thenA. ` 0.8 ` JB. ` 8 ` JC. `16 ` JD. ` 4 ` J |
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Answer» Correct Answer - B Stored energy ` U = ( 1 )/( 2 ) Li ^ 2 = ( 1 )/( 2 ) L ( ( E)/( R)) ^ 2 ` Here, `E = 10 V, L = 4 H and R = 5 Omega ` ` therefore U = ( 1 )/(2 ) xx 4 xx (( 10)/ ( 5 ))^ 2 J = 8 J ` |
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| 24. |
The instrument which works on the principle of mutual inductance isA. galvanometerB. ammeterC. potentiometerD. transformer |
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Answer» Correct Answer - D A transformer is based on the principle of mutual inductance. |
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| 25. |
What is the self inductance of a coil in which an induced emf of 2V is set up, when the current is changing at the rate of ` 4 A s ^( - 1 ) `A. 0.5 mHB. 0.05 HC. 2 HD. 0.5 H |
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Answer» Correct Answer - D Induced emf ` e = ( Ldi) /( dt )`, hence ` e = 2V, (di )/( dt ) = 4 As ^( -1 ) ` ` rArr 2 = L xx 4 ` ` L = ( 2 )/( 4 ) = 0. 5 H ` |
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| 26. |
A solide ball of volume V is dropped in a viscous liquid. It experiences a viscous force F. If the solid ball of volume 2V of same material is dropped in the same fluid, then the viscous force acting on it will beA. ` nF// 2 `B. `F// 2 `C. `2F `D. `2 nF ` |
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Answer» Correct Answer - C Viscous force , F = ` 6 pi eta r v ` but ` v prop r ^ 2 ` ` therefore F prop V " " ` ( volume ) so, on doubline the volume of the ball, the viscous force is doubled. |
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| 27. |
A ball of mass 0.1 kg strikes a wall normally with a speed of ` 30 m s ^( - 1 ) ` and rebounds with a speed of ` 20 ms ^( - 1 ) `. The impulse of the force exerted by the wall on the ball isA. ` 1 N - s `B. `5 N - s `C. `2 N-s `D. `3 N -s ` |
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Answer» Correct Answer - B Impulse = Change is momentum ` p _ i- P _ f ` direction of ` P _ f and p_i` apposite to each other. ` therefore ` Impulse = `m u - ( - mv ) ` ` = m u + mv = m ( u + v ) = 0.1 ( 30 + 20 ) ` ` = 0.1 xx 50 =5 N - s ` |
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| 28. |
An elevator is moving vertically up with a acceleration a the force exerted on the floor by a passenger of mass m isA. mgB. maC. mg - maD. mg + ma |
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Answer» Correct Answer - D `R = m ( g + a ) = mg + ma ` |
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| 29. |
Haber process is used for the production of which of the following ?A. ` NH _ 3 `B. ` HNO _ 3 `C. `H _ 2 SO _ 4 `D. ` O _3 ` |
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Answer» Correct Answer - A Haber process is used for the production of ammonia gas. ` N _ 2 (g ) + 3 H _2 (g ) hArr 2 NH _ 3 ( g ) ` |
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| 30. |
Gypsum isA. ` Ca SO _ 4 *H _ 2 O `B. `C aS O _ 4 * 2 H _ 2 O `C. `2 CaSO_ 4 * 2 H _ 2 O `D. `CaSO_4 ` |
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Answer» Correct Answer - B Gypsum is ` CaSO _ 4 * 2H _ 2 O `. |
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| 31. |
The minimum numbers of carbon atoms in ketones which will show chain isomerism will beA. 7B. 4C. 6D. 5 |
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Answer» Correct Answer - D There are minimum five carbon atoms in ketons which will show chain isomerism (i) ` CH _ 3 - underset (O ) underset ( "||")C - CH _ 2 - CH _ 2 - CH _ 3 ` (ii) ` CH _ 3 -underset (O ) underset ("||")C - underset ( CH_ 3 ) underset ( |) ( CH ) - CH _ 3 ` |
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| 32. |
Which of the following organic compounds could not be dried by anhydrous ` Ca Cl_ 2 ` ?A. EthanolB. BenzeneC. ChloroformD. Ethyl acetate |
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Answer» Correct Answer - A Ethanol could not be dried by anhydrous ` CaCl_ 2 ` because it reacts with `CaCl _2` as follows ` CaCl _ 2 + 4C _ 2 H _ 5 OH to underset ("solid") (CaCl _ 2 * 4 C _ 2 H _ 5 OH ) ` |
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| 33. |
C - H bond length is least inA. acetyleneB. methaneC. ethyleneD. ethane |
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Answer» Correct Answer - A C- H bond length is least for ` C-= ` C bonds. Thus, it is least in acetylene ` (CH -= CH) `. |
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| 34. |
The laughing gas isA. ` N _ 2 O _ 4 `B. `NO `C. `N _ 2 O `D. ` N _ 2 O _ 5 ` |
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Answer» Correct Answer - C Nitrous oxide , ` N _ 2 O ` is termed as laughing gas. |
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| 35. |
Which of the following is least hydrolysed ?A. ` Be Cl _ 2`B. `Mg Cl _ 2 `C. `CaCl _ 2 `D. `BaCl_ 2 ` |
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Answer» Correct Answer - A ` BeCl _ 2 ` is covalent whereas all other group - II metal halides are electrovalent. Therefore, ` BeCl _ 2 ` is least hydrolysed. |
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| 36. |
Which of the following statement is false ?A. 40 % solution HCHO is known as formalinB. HCHO is least reactive in its homologous seriesC. The boiling point of iso - vareladehyde less than n - varelaldehyde (4 )D. The boiling point of ketones are higher than that of aldehydes |
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Answer» Correct Answer - B Formaldehyde is the most reactive n its homologous series. Order of reactivity of some aldehydes is as follows. ` HCHO gt CH _ 3 CHO gt CH _ 3 CH_ 2 CHO ` |
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| 37. |
By which of the following process hydrocarbons are found from petroleum ?A. CombustionB. Fractional distillationC. AdditionD. All of the above |
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Answer» Correct Answer - B Hydrocarbons are distilled from petroleum by fractional distrillation method. |
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| 38. |
LPG mainly consist of the followingA. methaneB. hydrogenC. acetyleneD. butane |
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Answer» Correct Answer - D Liquified petroleum gas (LPG) mainly consist butane. Besides this, it also consists propane, methane, nitrogen and ethane in little amount. |
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| 39. |
Which of the following is a weak acid ?A. ` C _ 6 H _ 6 `B. `CH _ 3 COOH `C. `CH _ 2 = CH _ 2 `D. ` CH _ 3 COCH _ 3 ` |
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Answer» Correct Answer - A Due to delocalization of electrons in benzene , it is least acidic in nature. |
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| 40. |
Which of the following compound is used as a refrigerant ?A. ` C C l _ 4 `B. `C Cl _ 2 F _ 2 `C. `CF _4 `D. Acetone |
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Answer» Correct Answer - B Chloro fluoro carbons ` (C Cl _ 2 F _ 2 ) ` is used as regregerent. |
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| 41. |
Which of the following aromatic compound gives sulphonation reaction very easily ?A. ChlorobenzeneB. NitrobenzeneC. TolueneD. Benzene |
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Answer» Correct Answer - C Toluene gives sulphonation reaction very quickly due to H - bond formation. |
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| 42. |
The geometry of ` I _ 3 ^( - ) ` isA. triangularB. linearC. tetrahedralD. T - shape |
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Answer» Correct Answer - B ` I _ 3 ^ - ` ion is ` sp ^ 3 ` d - hybrisdised. Due to the presence of two lone pair of electrons its geometry beomes linear instead of trigonal bipyramidal. |
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| 43. |
The volume concentration of hydrogen peroxide having 6.8 % concentration will beA. 5B. 11.2C. 22.4D. 20 |
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Answer» Correct Answer - C Volume concentration ` = (112 )/( 34 ) xx % ` concentration ` = (112 ) /( 34 ) xx 6.8 = 22.4 ` |
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| 44. |
The molar concentration of chloride ions in the resulting solution of 300 mL of 3.0 M NaCl and 200 mL of 4.0 M ` BaCl _ 2 ` will beA. `1.7 ` MB. `1.8 ` MC. ` 5.0 ` MD. ` 3.5 ` M |
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Answer» Correct Answer - C ` underset (" 3.0 M") ( NaCl) hArr Na ^ + + underset("3.0 M")(Cl ^ - ) ` `underset ("4.0 M") (BaCl_2)hArr Ba^( 2 + ) + underset ( 2xx 4.0 M) ( 2Cl^-) ` ` therefore ` Molar concentration of ` Cl^- = ( M _ 1 V _ 1 + M _ 2 V _ 2 ) /( V _ 1 + V _ 2 ) ` ` = (3.0 xx 300 + 2 xx 4.0 xx 200 ) /( 300 + 200 ) ` ` = ( 900 + 1600 )/( 500 ) ` ` = 5.0 M ` |
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| 45. |
If ` x^(p) + y^(q) = (x + y)^(p+q) , " then" (dy)/(dx)` isA. ` - ( x ) /( y ) `B. ` ( x ) /( y ) `C. `- ( y ) / ( x ) `D. ` ( y ) /( x ) ` |
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Answer» Correct Answer - D if ` x ^p + y ^ q = ( x + y ) ^ ( p + q ) ` Taking log on both sides , ` p log x + q log y = ( p + q ) log ( x + y ) ` On differentiating w.r.t x, we get `( p ) /( x ) + ( q ) /( y ) * ( dy ) /( dx ) = (( p + q ))/( ( x + y)) ( 1 + (dy ) /(dx )) ` ` { (p) / ( x ) - ( p+ q )/( x + y )} = { ( p + q )/( x + y ) - ( q )/( y ) } (dy ) /( dx ) ` ` { ( px + py - px - qx ) /( x ( x + y ))} = { ( py + qy - qx - qy ) /( y ( x + y ))} (dy)/(dx ) ` ` rArr (( py - qx ) ) /( x ) = ((py- qx ) ) / (y ) * ( d y ) / ( dx ) ` `rArr ( dy)/ (dx ) = (y )/ ( x ) ` |
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| 46. |
All points on the curve `y^(2)=4a(x+a" sin"(x)/(a))` at which the tangents are parallel to the axis of x lie on aA. circleB. parabolaC. straight lineD. None of these |
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Answer» Correct Answer - B ` y ^ 2 = 4a [ x + a sin (( x ) /( a ))]" " `…(i) ` therefore 2y ( dy )/ ( dx ) = 4a [ 1 + cos ( ( x )/(a )) ]" " `… (ii) If tangent is parallel to x - axis , then ` (dy ) /( dx ) = 0 ` So, from Eq. (i), we get ` cos (( x )/( a )) = - 1 ` ` therefore sin (( x )/( a ))= 0 ` On putting this value in Eq. (i) , we get ` y^ 2 = 4a ( x + 0 ) rArr y ^ 2 = 4a x ` So, all the points on the curve ` y^ 2 = 4a ( x + a sin "" ( x ) /( a )) `, where the tangents is parallel to the x-axis are lies on parabola. |
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| 47. |
The equation of directrix is to the parabola ` 4x ^ 2 - 4x - 2y + 3 = 0 ` will beA. `2y = 1 `B. ` 2x = 1 `C. `2y = 3 `D. ` 2x = 3 ` |
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Answer» Correct Answer - A Given parabola , ` 4x ^ 2 - 4x - 2y + 3= 0 ` ` rArr 4 ( x ^ 2 - x ) = 2y - 3 ` ` rArr 4 ( x ^ 2 - x + ( 1 ) /( 4 ) - ( 1 )/( 4 )) = 2y - 3 ` ` rArr ( x - ( 1 ) /( 2 ) ) ^ 2 = 2y - 2 ` ` rArr ( x - (1 ) /(2 ) ) ^ 2 = 2 ( y -1 ) ` Which is of form ` X^ 2 = 4aY ` where , ` X = x - (1 ) / ( 2 ) and Y = y - 1 ` and ` a = (1 )/ ( 2 ) ` ` therefore ` Directrix, ` Y + a = 0 ` ` rArr y - 1 + ( 1 ) / ( 2 ) = 0 ` ` rArr y - ( 1 ) / ( 2 ) = 0 ` ` rArr 2y = 1 ` |
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| 48. |
The length of normal at any point to the curve, `y=c cosh(x/c)` isA. fixedB. ` ( y^ 2 )/( c^ 2 ) `C. ` (y ^ 2 ) /( c ) `D. ` ( y ) /( c ^ 2 ) ` |
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Answer» Correct Answer - C Given `, y = c cos h (( x)/ ( c )) ` ` ( d y ) /( dx ) = c * ( 1 )/ (c ) * sin h (( x )/ ( c )) = sin h ( ( x )/( c )) ` Now, length of normal ` = y sqrt ( 1 + (( dy )/( dx ))^ 2 ) ` ` = c cos h ( ( x )/( c )) sqrt ( 1 + sin ^ 2 h (( x )/ (c )) ) ` ` = c cos h (( x )/ (c )) sqrt ( cos h^ 2 ( ( x )/( c )) ) ` ` = c[ cos h (( x )/ ( c))]^ 2 ` ` = c (( y ) / ( c ))^ 2 " " `[from Eq. (i) ] ` = ( y^ 2 )/( c ) ` |
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| 49. |
If ` tan x = (2t)/(1 -t^(2)) " and sin y" = (2t)/(1 + t^(2))` , then the value of ` (dy)/(dx)` isA. 1B. tC. ` ( 1 ) /( 1 - t ) `D. ` ( 1 ) /( 1 + t ) ` |
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Answer» Correct Answer - A Given, ` tan x = ( 2t )/( 1- t ^ 2 ) and sin y = ( 2t ) /( 1 + t ^ 2 ) ` Now,` x =tan ^( -1 ) ( ( 2t ) /( 1- t^ 2)) ` ` x = 2 tan ^( - 1 ) t " " `… (i) and ` y = sin ^( - 1 ) ((2t ) /( 1 + t^2 )) ` ` y = 2 tan ^( -1 ) t " " `... (ii) From Eq. (i) ` (dx )/ ( dt ) = ( 2 )/( 1 +t ^ 2 ) ` From Eq. (ii) ` ( dy )/( dt ) = ( 2 )/( 1 + t^ 2 ) ` ` therefore (dy )/(dx) = ( dy ) /(dt ) xx ( dt ) /( dx ) = ( 2 ) /( (1 + t^ 2 )) xx (( 1 + t^ 2 ))/( 2 ) = 1 ` |
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| 50. |
The line `lx+my+n=0` is a normal to the parabola `y^2 = 4ax` ifA. `mn = al ^ 2 `B. ` lm = an ^ 2 `C. `ln = am ^ 2 `D. None of the above |
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Answer» Correct Answer - C Given ,parabola, ` y ^ 2 = 4 ax ` ` rArr 2y ( dy ) /( dx ) = 4a ` ` rArr ( d y ) /( dx ) = (2a ) /( y ) " " ` … (i) which is the slope of tangent Given ` lx + my + n = 0 ` is an equation of tangent of the parabola `y^ 2 =4a x ` ` therefore ` Slope of tangent ` = - ( 1 ) /( m ) " "` ... (ii) From Eqs (i) and (ii) ` (2a )/( y ) = - ( l ) / ( m ) rArr y = ( - 2am ) /( l ) ` ` because y ^ 2 = 4ax ` ` rArr ( 4a ^ 2 m^ 2 ) /( l ^ 2 ) = 4ax` ` rArr x = ( am^ 2 ) /( l^ 2 ) ` On putting the values of x and y in the following equation ` l ( ( am^2 ) /( l^2 )) +m ( ( -2 am ) /( l ) ) + n = 0` ` ( am ^ 2 ) /( l ) -(2am ^ 2 ) /( l) + n = 0 ` ` rArr ( am^ 2 ) / ( l) = n rArr am ^ 2 = n l` Which is the required relation. |
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