Ask & Answer: India's Largest Education Community

Kindly go through the solution

This is a short answer type question as classified in NCERT Exemplar

V $\propto T$

V/T = constant

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

T₁=273+27=300K

T₂= 273+327= 600K

V₁= 100cc

V₂=V₁ (600/300)

V₂=2V₁

V₂= 2 (100)=200cc

The primary purpose of Figures of Speech is to be more creative and expressive with the language or communication. It also improves the reader experience. Figures of Speech are also to enhance and elevate the language. Moreover, it also adds depth and clarity for the readers. Last but not the least, Figures of Speech adds a touch of humour and comic relief and at times a dramatic effect, which are important components to garner and retain readers' attention.

Candidates must reach the allotted centre as per the date and time allotted in their UP CPET 2025 Admit Card. Any digital format (on the mobile) of the UP CPET admit card will not be accepted at the examination centre. Hence students are advised to bring the hard copy to the exam centre. 

Putting e^x = t so that e^x dx = dt=>dx = $\frac{dt}{e^x}=\frac{dt}{t}$

$\begin{array}{l}\therefore {\displaystyle \int \frac{dx}{e^x-1}={\displaystyle \int \frac{dt}{(t-1)\times t}={\displaystyle \int \frac{dt}{t(t-1)}}}}\\ ={\displaystyle \int \frac{t-(t-1)}{t(t-1)}}dt\\ ={\displaystyle \int \frac{dt}{t-1}}-{\displaystyle \int \frac{dt}{t}}\\ =log\left|t-1\right|-log\left|t\right|+\; C\\ =log\left|e^n-1\right|-log{e}^x+C.\end{array}$

The total number of seat intakes for the BBA course at Siva Sivani Degree College Hyderabad (SSDCH) is 120. To get a seat in this course, candidates are required to perform well in their Class 12. These 120 seats are divided equally among its offered specialsiations. Check out the number of seats allocated to each specialisation in the table below:

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- I=AneV_d  , current is directly proportional to drift velocity.

No, SAT used to stand for Scholastic Aptitude Test. The SAT acronym does not hold any meaning after becoming digital and the entire exam format witnessing a major change, the SAT exam full form is no longer valid and not used by College Board as well. SAT exam is addressed as SAT exam rather than using its full form. 

Putting x⁴ = tÞ 4x³dx = dt

$\begin{array}{l}\therefore {\displaystyle \int \frac{1}{x(x^4-1)}dx=\frac{1}{4}{\displaystyle \int \frac{dt}{t(t-1)}=\frac{1}{4}{\displaystyle \int \frac{t-(t-1)}{t(t-1)}}dt}}\\ =\frac{1}{4}\left\{{\displaystyle \int \frac{dt}{t-1}-{\displaystyle \int \frac{dt}{t}}}\right\}\\ =\frac{1}{4}\left\{log\left|t-1\right|-logt\right\}+\; C\end{array}$

$\begin{array}{l}=\frac{1}{4}log\left|\frac{t-1}{t}\right|+\; C\\ =\frac{1}{4}log\left|\frac{x^4-1}{x^4}\right|+\; C\end{array}$

There are a number of Figures of Speech which are similar to each other but have a subtle difference. Students often get confused between the two, but the confusion and common errors can be mitigated by understanding the differences and regular practice. Some such Figures of Speech are as follows:

·       Simile, Metaphor and Personification

·       Oxymoron and Paradox

·       Paradox and Antithesis

·       Hyperbole and Understatement

·       Apostrophe and Invocation.

Putting x² = t such that 2xdx = dt

$\begin{array}{l}{\displaystyle \int \frac{2xdx}{(x^2+1)(x^2+3)}={\displaystyle \int \frac{dt}{(t+1)(t+3)}=\frac{1}{2}{\displaystyle \int \frac{2dt}{(t+1)(t+3)}}}}\\ =\frac{1}{2}{\displaystyle \int \frac{(t+3)-(t+1)dt}{(t+1)(t+3)}}\\ =\frac{1}{2}\left\{{\displaystyle \int \frac{dt}{t+1}}-\frac{dt}{t+3}\right\}\\ =\frac{1}{2}\left\{log\left|t+1\right|-log\left|t+3\right|\right\}+\; c\\ =\frac{1}{2}log\left|\frac{t+1}{t+3}\right|+\; c\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- R= $\frac{\delta l}{A}$ for greater value of R, A should be less and its possible value when connected across 1cm $\times$ 1/2cm faces.

Kindly go through the solution

This is a short answer type question as classified in NCERT Exemplar

molar mass = mass of avogadro’s number of atoms= 6.023 $\times {10}^{23}$ atoms.

197 g of gold contains =6.023 $\times {10}^{23}atoms$

1g of gold contain= $\frac{6.023\times {10}^{23}}{197}$ atoms

39.4 g of gold atoms = $\frac{6.023\times {10}^{23}\times 39.4}{197}=1.2\times {10}^{23}$ atoms

There are broadly over 20 types of Figures of Speech. While around 12 are commonly used in day-to-day communication, rest are used specifically for literature elements such as poems, drama or plays, etc.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation-The potential drop along the wires of potentiometer should be greater than emfs of cells.
In a potentiometer experiment, the emf of a cell can be measured if the potential drop along the potentiometer wire is more than the emf of the cell to be determined. Here, values of emfs of two cells are given as 5 V and 10 V, therefore, the potential drop along the potentiometer wire must be more than 10 V.

Putting sin x = t cos xdx = dt

$\begin{array}{l}{\displaystyle \int \frac{dt}{(1-t)(2-t)}}={\displaystyle \int \frac{(2-t)-(1-t)}{(1-t)(2-t)}}dt.\\ ={\displaystyle \int \frac{dt}{1-t}-{\displaystyle \int \frac{dt}{2-t}}}\\ =\frac{log\left|1-t\right|}{(-1)}-\frac{log\left|2-t\right|}{(-1)}+c\\ =log\left|\frac{2-t}{1-t}\right|+c\\ =log\left|\frac{2-sinx}{1-sinx}\right|+c\end{array}$