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# Tutor profile: Utkarsh B.

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Utkarsh B.
Tutor for 4 years ,IChO Scholar
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## Questions

### Subject:Basic Math

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Question:

Find the range of values of t for which {\displaystyle 2\sin t={\frac {1-2x+5x^{2}}{3x^{2}-2x-1}}} {\displaystyle 2\sin t={\frac {1-2x+5x^{2}}{3x^{2}-2x-1}}}

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Utkarsh B.

Given 2sint=1−2x+5x23x2−2x−1⇒6sint⋅x2−4sint⋅x−2sint=1−2x+5x22sin⁡t=1−2x+5x23x2−2x−1⇒6sin⁡t⋅x2−4sin⁡t⋅x−2sin⁡t=1−2x+5x2 ⇒(6sint−5)x2+2(1−2sint)x−(2sint+1)=0⇒(6sin⁡t−5)x2+2(1−2sin⁡t)x−(2sin⁡t+1)=0 , Now for calculation of value of 2sint,2sin⁡t, equation must have real roots. So D≥0. So 4(1−2sint)2+4(2sint+1)⋅(6sint−5)≥04(1−2sin⁡t)2+4(2sin⁡t+1)⋅(6sin⁡t−5)≥0 (1−2sint)2+(2sint+1)⋅(6sint−5)≥0(1−2sin⁡t)2+(2sin⁡t+1)⋅(6sin⁡t−5)≥0 So we get 16sin2t−8sint−4≥0⇒4sin2t−2sint−1≥0 now your equation reads as sintsin⁡t is in [−1,1−5√4]∪[1+5√4,1].[−1,1−54]∪[1+54,1]. the solution to this equation is what i claimed at the beginning of this answer. you can establish (a) by turning k=1−2x+5x23x2−2x−1k=1−2x+5x23x2−2x−1 into a quadratic equation (3k−5)x2−2(k−1)x−(k+1)=0(3k−5)x2−2(k−1)x−(k+1)=0 whose discriminant is 4(k2−k−1)4(k2−k−1) which is positive for kk in the range (−∞,1−5√2]∪[1+5√2,∞).

### Subject:Chemistry

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Question:

(i) A mixture of ideal gases is cooled upto liquid helium temperature (4.22 K) to form an ideal solution. Is this statement true or false ? Justifiy your answer in not more than two lines. (ii) Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen. (iii) The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at 25󿼇C are 3.0x 10-4 s-1, 104.4 kJ mo1-1 and 6.0 x 1014s -1 respectively. The value of the rate constant as T → &infin is: (a)2.0 x 1018s -1 (b)6.0 x 1014s -1 (c)Infinity (d)3.6 x 1030s -1 (iv) The orbital angular momentum of an electron in 2s orbital is: (a) {\displaystyle {\frac {1}{2}}{\frac {h}{2\pi }}} {\displaystyle {\frac {1}{2}}{\frac {h}{2\pi }}} (b) {\displaystyle {\frac {h}{2\pi }}} {\displaystyle {\frac {h}{2\pi }}} (c)Zero (d) {\displaystyle {\sqrt {2}}{\frac {h}{2\pi }}} {\displaystyle {\sqrt {2}}{\frac {h}{2\pi }}}

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Utkarsh B.

Ideal Gas Mixtures For a student who missed the tricky part of this question, it is going to be very confusing, because he would be left wondering if there is anything that has to be taken upto liquid helium temperature to be liquefied. The point here is that ideal gases are assumed to have zero intermolecular forces. No matter what the conditions are, they just do not have any IMF whatsoever to be liquefied! Check back the assumptions for ideal gas. Balmer Series The Balmer series corresponds to the energy level transition from some orbit to n=2. The shortest wavelength would be the one with the most energetic transition - naturally, n  ∞ {\displaystyle v=R_{H}\left({\frac {1}{2^{2}}}-{\frac {1}{\infty ^{2}}}\right)={\frac {R_{H}}{4}}} {\displaystyle v=R_{H}\left({\frac {1}{2^{2}}}-{\frac {1}{\infty ^{2}}}\right)={\frac {R_{H}}{4}}} Rate Constant Just recall the Arrhenius equation. K = A(e)^(-Ea/RT) When T tends towards infinity, the exponential term will tend towards 1, giving the value of the rate constant at infinite temperature = A. Ratio of RMS velocities Simple application of RMS = {\displaystyle {\sqrt {\frac {3RT}{M}}}} {\displaystyle {\sqrt {\frac {3RT}{M}}}} Taking the ratio, The required answer is {\displaystyle {\sqrt {\frac {M_{\left(oxygen\right)}T_{\left(hydrogen\right)}}{T_{\left(oxygen\right)}M_{\left(hydrogen\right)}}}}=1} {\displaystyle {\sqrt {\frac {M_{\left(oxygen\right)}T_{\left(hydrogen\right)}}{T_{\left(oxygen\right)}M_{\left(hydrogen\right)}}}}=1} Note By the way, most people, including relatively better placed students; make a little mistake of substituting molar masses in gm/mol in the equation of RMS speed. Whether or not you convert it into Kg/Mol, it will not matter here (because its a ratio that we calculated) but will introduce a BIG error (if at all you can call it an error and not a mistake!) in the calculation of the absolute value. Orbital Angular Momentum This is a trivial piece of information. The orbital angular momentum is {\displaystyle {\frac {{\sqrt {\left(l\right)}}\left(l+1\right)h}{2\pi }}} {\displaystyle {\frac {{\sqrt {\left(l\right)}}\left(l+1\right)h}{2\pi }}}. (nh/2π) is the angular momentum.

### Subject:Physics

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Question:

On a boat, there is a 15 inch brick of gold and a 15 inch brick of iron. If both of them are dropped into the water, which one of them will make the water level higher?

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Utkarsh B.

The iron brick will make the water level higher. Explanation: But on the contrary, in actual terms, none of them will help in rising the water. They will actually drop the water level. This is because, they have a large effect on the water level while in the boat but when they are dropped into the water. They will drop the water level. Iron brick being the lighter one will drop the water level less because it will have less effect on the water from the boat but will displace the equal amount of water as the gold when it is dropped into the water. Thus it can be said that it will make the water level higher.

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