An alpha active element has molar mass M and has a half life of `tau`. A large plate of thickness `alpha` is made from the radioactive element. Density of the plate is `rho`. Calculate the number of alpha particles emitted in unit time from unit surface area of the plate. Avogadro’s number is `N_(A)`.
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A radioactive element has half-life period of 30 days. How much of it will be left after 90 days?
Correct Answer - C `90 "days" rarr 3 "half lives"`, left `1/8` i.e., `12.5%` Disintegrated `= 100-12.5 = 87.5%`
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The half life a radioactive element is `30` days, if the mass of the specimen reduces to `1//10^(th)` then the time taken is :-
Correct Answer - C `N = N_(0)e^(-lambdat) rArr (N_(0))/(10) = N_(0)e^(-lambdat) rArr ln(1) - ln(10) = - lambdat` `rArr t = ((ln10)/(ln2)) T_(1//2) rArr t = 33` days
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If a variable takes the discrete values `alpha-4`, `alpha -(7)/(2), alpha-(5)/(2), alpha-2,alpha+(1)/(2), alpha-(1)/(2), alpha+5(alpha gt 0)`, then th
Correct Answer - A Arrange the data as follows : `alpha-(7)/(2), alpha-3, alpha-(5)/(2),alpha-2, alpha-(1)/(2),alpha+(1)/(2),alpha+4,alpha+5` Median `=(1)/(2)` [value of 4th item+value of 5th item] `therefore " Median"=(alpha-2+alpha-(1)/(2))/(2)=(2alpha-(5)/(2))/(2)=alpha-(5)/(4)`
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If `A(alpha, beta)=[("cos" alpha,sin alpha,0),(-sin alpha,cos alpha,0),(0,0,e^(beta))]`, then `A(alpha, beta)^(-1)` is equal to
Correct Answer - A We have, `A(alpha, beta)^(-1)=1/e^(beta) [(e^(beta) cos alpha,-e^(beta) sin alpha,0),(e^(beta) sin alpha,e^(beta) cos alpha,0),(0,0,1)]` `=A(-alpha, -beta)`
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There are two sets of parallel lines, their equations being `x cos alpha+y sin alpha=p` and `x sin alpha- y cos alpha=p` , `p=1,2,3,….n` and `alpha in
Correct Answer - C `(c )` `"^(n)C_(2)*^(n)C_(2)=225impliesn=6`
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You have to design a half-wave plate of quartz crystal for light of wavelength 5800Å. Calculate the thickness of the plate rquired for the purpose if
Correct Answer - `3.2xx10^(-5)`m
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A parallel - plate capacitor of plate area `A` and plate separation d is charged to a potential difference `V` and then the battery is disconnected .
Correct Answer - decrease in energy is `(1)/(2)(epsilon_(0)AV^(2))/(d)((epsilon_(r)-1)/(epsilon_(r)));` energy is transferred to the agent introducing the slab
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A parallel-plate capacitor contains a mica sheet (thickness `10^(-3)`m) and a sheet of fibre (thickness `0.5xx10^(-3)`m). The dielectric constant of m
Correct Answer - `5200 V`
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A metal plate with a total surface are of `300cm^(2)` is to be benickel plate. If a current of 1.5A is used for 3 hours, find the thickness of nickel
Correct Answer - `1.81xx10^(-5)m`
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Between the plates of a parallel-plate capacitor there is a metallic plate whose thickness takes up `eta = 0.60` of the capacitor gap. When that plate
Correct Answer - (a)`-0.15mJ` (b) `0.15mJ`
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