A girl throws a ball vertically upwards.It takes a different time intervals \(\) of \(3.2s\) to return to her hand.
Assume air resistance is negligible .What is the intial speed with which the ball is thrown?
(A) \(3.07ms^{-1}\)
(B) \(7.85ms^{-1}\)
(C) \(15.7 ms^{-1}\)
(D) \(31.4 ms^{-1}\)
\(S=ut_1-\frac {1} {2} gt_1^2\)
\(=ut_2-\frac {1} {2} gt_2^2\)
\(t_1+t_2=3.2\)
since \(t_1\ne t_2\)
\(u(t_1-t_2)=\frac {1} {2} g(t_1^2-t_2^2)\)
\(u=\frac {1} {2}\times 3.2\times g\)
\(u=15.7ms^{-1}\)
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This Question is about gases
Figure 3 shows how nitrogen is used in the Haber Process to Produce ammonia
Gas X in figure 3 Obtained from methane.
Name gas X _______________
(A) \(O_2\)
(B) \(H_2\)
(C) \(N_2\)
(D) \(CO_2\)
\(H_2\)
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Given that \(f(x)=5x^4-3+\frac {2x^2-x^{\frac {3} {2}}} {\sqrt {x}},x>0\)
Where \(\frac {2x^2-x^{\frac {3} {2}}} {\sqrt {x}}\) is in the form \(2x^p-x^q\). Calculate \(f'(0)\)
(A) \(-2\)
(B) \(\frac {3} {2}\)
(C) \(-1\)
(D) \(\frac {1} {2}\)
Since \(\frac {2x^2-x^{\frac {3} {2}}} {\sqrt {x}}\)can be written as
\(=\frac {2x^2} {x^{\frac {1} {2}}}-\frac {x^{\frac {3} {2}}} {x^{\frac {1} {2}}}\)
\(=2x^{\frac {3} {2}} -x^1\)
\(f(x)=5x^4-3+2x^{\frac {3} {2}}-x\)
\(f'(x)=20x^3+3x^{\frac {1} {2}}-1\)
\(f'(0)=-1\)
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A curve has equation \(y=2x^3-2x^2-2x+8\)
At which value of positive integral value of x is \(f'(x)=3.f''(x)-14\)
(A) \(x=3\)
(B) \(x=9\)
(C) \(x=6\)
(D) \(x=12\)
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\((4+3\sqrt{x})^2\) can be written as \(16+k\sqrt{x}+9x,\)where k is a constant.What is approximate value of \(4+3\sqrt{k}\) ?
(A) 18.696
(B) 12.966
(C) 14.336
(D) 24.112
K=24
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If \(\alpha\) and \(\beta\) be the coefficients of \(x^4\) and \(x^2\) respectively in the expansion of
\((x+\sqrt{x^2-1}\big)^6+\big(x-\sqrt{x^2-1}\big)^6, \) then
(1) \(\alpha+\beta=60\)
(2) \(\alpha+\beta=-30\)
(3) \(\alpha-\beta=-132\)
(4) \(\alpha-\beta=60\)
\(2[^6C_0x^6+^6C_2x^4(x^2-1)+^6C_4x^2(x^2-1)^2+^6C_6(x^2-1)^3]\)
\(\alpha=-96\ \&\ \beta=36\)
\(\therefore \alpha-\beta=-132\)
(3) Option
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\(\displaystyle \lim_{x \to \frac{\pi}{2}}\frac{\cot x-\cos x}{(\pi-2\pi)^3}\) equals
(1) \(\frac{1}{8}\) (2) \(\frac{1}{4}\)
(3) \(\frac{1}{24}\) (4) \(\frac{1}{16}\)
\(\displaystyle \lim_{x \to \frac{\pi}{2}}\frac{\cot x-\cos x}{(\pi-2x)^3}\)
Put, \(\frac{\pi}{2}-x=t\)
\(\displaystyle \lim_{t \to 0}\frac{\tan t-\sin t}{8t^3}\)
\(=\displaystyle \lim_{t \to 0}\frac{\sin t\cdot 2\sin^2\frac{t}{2}}{8t^3}\)
\(=\frac{1}{16}.\)
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General solution of differential equation \(\frac{dy}{dx}+y=1(y\neq1)\) is
(1) \(\log \Bigg|\frac{1}{1-y}\Bigg|=x+C\)
(2) \(\log\big|1-y\big|=x+C\)
(3) \(\log \big|1+y\big|=x+C\)
(4) \(\log\Bigg|\frac{1}{1-y}\Bigg|=-x+C\)
We have,
\(\frac{dy}{dx}+y=1\)
\(\Rightarrow \frac{dy}{dx}=1-y\)
\(\Rightarrow\frac{1}{1-y}dy=dx\)
On integrating both sides, we have
\(\int \frac{1}{1-y}dy=\int dx\)
\(\Rightarrow -\log|1-y|=x+C\)
\(\Rightarrow \log|\frac{1}{1-y}|=x+C\)
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The value of numerical aperature of the objective lens of a microscope is 25. If light of wavelength 5000 Å is used, the minimum separation between two points, to be seen as distinct, will be :
(A) 0.48 \(\mu m\)
(B) 0.38 \(\mu m\)
(C) 0.24 \(\mu m\)
(D) 0.12\(\mu m\)
Numerical aperature of the microscope is given as
\(NA=\frac {0.61\lambda} {d}\)
Where d = minimum separaton between two points to be seen as distinct
\(d=\frac {0.61\lambda} {NA}=\frac {(0.61)\times(5000\times10m^{-10})}{1.25}\\=2.4\times10^{-7}m\\=0.24\mu m\)
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