A coil of radius R carries current `i_1`. Another concentric coil of radius `r(rlt ltR)` carries current `i_2`. Planes of two coils are mutually perpendicular and both the coil are free to rotate about common diametre. Find maximum kinetic energy of smaller coil when both the coils are released, masses of coils are M and m, respectively.
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Find the mutual inductance of two concentric coils of radii `a_(1)` and `a_(2)(a_(1) lt lt a_(2))` if the planes of the coils make an angle `theta` wi
If `i` current flows in the larger coil, magnetic field produced at the centre will be perpendicular to the plane of larger coil. Now the area vector of smaller coil...
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`IfI_1=int_0^1 2x ,I_2=int_0^1 2^x^3dx ,I_3=int_1^22^x^2dx ,I_4=int_1^2 2^x^3dx ,` then which of the following is/are ture? `I_1> I_2` (b) `I_2> I_2`
Correct Answer - A::D For `0lt xlt 1, x^(3)gtx^(3)` or `2^(x^(2))gt2^(x^(3))` or `int_(0)^(1)2^(x^(2))dxgt int_(0)^(1)2^(x^(3))dx` Hence `I_(1)gtI_(2)` Also for `1ltxlt2, x^(2)ltx^(3)` or `2^(x^(2))lt2^(x^(3))` or `int_(1)^(2)2^(x^(2))dxlt int_(1)^(2)2^(x^(3))dx` or `I_(3)ltI_(4)`
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`IfI_I=int_0^(pi//2)cos(sinx)dx ,I_2=int_0^(pi/2)sin(cosx)d ,a n dI_3=int_0^(pi/2)cosx dx ,` then find the order in which the values `I_1,I_2,I_3,` ex
Correct Answer - `I_(1)gtI_(3)gtI_(2)` We know that `cosx` is decreasing function in `(0,pi//2)`. Also `xgtsinx` for `xepsilon(0,pi//2)` Thus, `cosx lt cos (sinx)` Further, `xgtsinx` and `cosxepsilon(0,1)` for `xepsilon(0,pi//2)` `:.cosxgtsin(cosx)` Thus, `sin...
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If `I_1=int_0^pixf(sin^3x+cos^2x)dxa n d` `I_2=int_0^(pi/2)f(sin^3x+cos^2x)dx ,t h e nr e l a t eI_1a n dI_2`
Correct Answer - `I_(1)=piI_(2)` `I_(1)=int_(0)^(1)(pi-x)f(sin^(3)x+cos^(2)x)dx` `:. 2I=pi int_(0)^(pi)f(sin^(3)x+cos^(2)x) dx` `=2pi int_(0)^(pi//2) f(sin^(3)x+cos^(2)x)dx` or `I_(1)=pi int_(0)^(pi//2) f(sin^(3)x+cos^(2)x)dx=piI_(2)`
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The numbers of possible continuous `f(x)` defined in `[0,1]` for which `I_1=int_0^1f(x)dx=1,I_2=int_0^1xf(x)dx-a ,I_3=int_0^1x^2f(x)dx=a^2i s//a r e`
Correct Answer - D Since `a^(2)I_(1)-2aI_(2)+I_(3)=0` `int_(0)^(1)(a-x)^(2)f(x)dx=0` Hence, there is no such positive function `f(x)`.
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`"I f"I_1=int_(-100)^(101)(dx)/((5+2x-2x^2)(1+e^(2-4x)))` `"and"I_2=int_(-100)^(101)(dx)/(5+2x-2x^2),t h e n(I_1)/(I_2)"i s"` 2 (b) `1/2` (c) 1 (d) `-
Correct Answer - B `I_(1)=int_(-100)^(101)(dx)/((52x-2x^(2))(1+e^(2-4x)))` `=int_(-100)^(101)(dx)/((5+2(1-x)-2(1-x)^(2)(1+e^(2-4(1-x))))` or `2I_(1)=int_(-100)^(101)(dx)/(5+2x-2x^(2))=I_(2)` or `(I_(1))/(I_(2))=1/2`
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`Iff(x)=(e^x)/(1+e^x),I_1=int_(f(-a))^(f(a))xg(x(1-x)dx ,a n d` `I_2=int_(f(-a))^(f(a))g(x(1-x))dx ,t h e nt h ev a l u eof(I_2)/(I_1)i s` `-1` (b) `-
Correct Answer - C `f(x)=(e^(x))/(1+e^(x))` `:.f(a)=(e^(a))/(1+e^(a))` and `f(-a)=(e^(-a))/(1+e^(-a))=(e^(-a))/(1+1/(e^(a)))=1/(1+e^(a))` `:. f(a)+f(-a)=(e^(a)+1)/(1+e^(a))=1` Let `f(-a)=alpha` or `f(a)=1-alpha` Now `I_(1)=int_(alpha)^(1-alpha)xg(x(1-x))dx` `=int_(alpha)^(1-alpha)(1-x)g((1-x)(1-(1-x))dx` `=int_(alpha)^(1-alpha) (1-x)g(x(1-x))dx` `:. 2I_(1)=int_(alpha)^(1-alpha) g(x(1-x))dx=I_(2)` or `(I_(2))/(I_(1))=2`
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`IfI_1=int_0^(pi/2)(cos^2x)/(1+cos^2x)dx ,I_2=int_0^(pi/2)(sin^2x)/(1+sin^2x)dx` `I_3=int_0^(pi/2)(1+2cos^2xsin^2x)/(4+2cos^2xsin^2x)dx ,t h e n` `I_1
Correct Answer - C `I_(1)int_(0)^(pi//2)(cos^(2)x)/(1+cos^(2)x)dx` `=int_(0)^(pi//2)(cos^(2)(pi//2-x))/(1+cos^(2)(pi//2-x))dx` `=int_(0)^(pi//2) (sin^(2)x)/(1+sin^(2)x)dx=I_(2)` Also `I_(1)+I_(2)=int_(0)^(pi//2)((sin^(2)x)/(1+sin^(2)x)+(cos^(2)x)/(1+cos^(2)x))dx` `=int_(0)^(pi//2)(sin^(2)x+sin^(2)xcos^(2)x+cos^(2)x+sin^(2)xcos^(2)x)/(1+sin^(2)x+cos^(2)x+sin^(2)xcos^(2)x)` `=int(1+2sin^(2)xcos^(2)x)/(2+sin^(2)x cos^(2)x)dx=2I_(3)` `:. 2I_(1)=2I_(3)` or `I_(1)=I_(3)` or `I_(1)=I_(2)=I_(3)`
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If `I_1=int_0^1(e^x)/(1+x) dx`aand `I_2=int_0^1 x^2/(e^(x^3)(2-x^3)) dx` then `I_1/I_2` is
Correct Answer - C `I_(1)=int_(0)^(1)(e^(x)dx)/(1+x), I_(2)=int_(0)^(1)(x^(2)dx)/(e^(x^(2))(2-x^(3)))` In `I_(2)` put `1-x^(3)=t` `:. I_(2)=1/3 int_(1)^(0) (-dt)/(e^(1-t)(1+t))=1/(3e)int_(0)^(1)(e^(t)dt)/(1+t)=1/(3e)I_(1)` `:. (I_(1))/(I_(2))=3e`
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If in triangle `A B C ,sumsinA/2=6/5a n dsumI I_1=9` (where `I_1,I_2a n dI_3` are excenters and `I` is incenter, then circumradius `R` is equal to `(1
Correct Answer - A We know that `II_(1) = 4R sin A//2` Now, `sum II_(1) = 9` `rArr 4R sum sin.(A)/(2) = 9` or `4R ((6)/(5)) = 9 " or "...
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