Thermal Conductivity is defined as the quantity of heat (Q) transmitted through a unit thickness (l) in a direction normal to a surface of unit area due to a unit temperature gradient (ΔT) under steady state condition and when the heat transfer is dependent only on the temperature gradient.

From Ohm’s Law, if current (I) flows through a resistance at potential difference (V₁ – V₂), then (V₁ – V₂) = IR (Or) \(I=\frac{\left( {{V}_{1}}-{{V}_{2}} \right)}{R}\).

Electrical Resistance \(\left( R \right)=\frac{\rho l}{A}=\frac{l}{\sigma A}\).

Where,

R = Resistance of conductor,

l = Length of conductor,

σ = Conductivity of the material.

\(Current\left( I \right)=\frac{\Delta Q}{\Delta t}=\frac{\left( {{V}_{1}}-{{V}_{2}} \right)}{\left( \frac{l}{\sigma A} \right)}\).

\(Current\left( I \right)=\frac{\Delta Q}{\Delta t}=\frac{\sigma A\left( {{V}_{1}}-{{V}_{2}} \right)}{l}\) … (1)

If a metal rod of length (l), cross – section area (A) and thermal conductivity (K) connects two bodies at temperature θ₁ and θ₂ (θ₁ > θ₂). The rate of heat flow or heat current:

\(H=\frac{\Delta Q}{\Delta t}=\frac{\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}{{{R}_{T}}}\).

Thermal resistance of the rod:

\({{R}_{T}}=\frac{l}{KA}\).

∴ \(H=\frac{\Delta Q}{\Delta t}=\frac{\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}{\left( \frac{l}{KA} \right)}=\frac{KA\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}{l}\) … (2)

Equation (2) is mathematically equivalent to Equation (1), with temperature. Hence, results derived from Ohm’s law are also valid for thermal conduction.

]]>Let n is a rational number and x is a real number such that |x| < 1, then \({{\left( 1+x \right)}^{n}}=1+nx+\frac{n(n-1)}{2!}{{x}^{2}}+\frac{n(n-1)(n-2)}{3!}{{x}^{3}}+……\).

(i) The
expansion of (x + a)^{n}

\(={{a}^{n}}{{\left( 1+\frac{x}{a} \right)}^{n}}\).

\(={{a}^{n}}\left( 1+n\times \left( \frac{x}{a} \right)+\frac{n(n-1)}{2!}{{\left( \frac{x}{a} \right)}^{2}}+\frac{n(n-1)(n-2)}{3!}{{\left( \frac{x}{a} \right)}^{3}}+….. \right)\).

The above expansion is valid only, when |x/a|< a.

(ii) The expansion (2 + 3x)⁻⁵ up to four terms in decreasing power of x is as follows.

\({{(2+3x)}^{-5}}={{\left[ 3x\left[ 1+\frac{2}{3x} \right] \right]}^{-5}}\).

\(=\frac{1}{243{{x}^{5}}}\left[ 1+(-5)\left( \frac{2}{3x} \right)+\frac{(-5)(-5-1)}{2!}{{\left( \frac{2}{3x} \right)}^{2}}+\frac{(-5)(-5-1)(-5-2)}{3!}{{\left( \frac{2}{3x} \right)}^{3}}+…. \right]\).

\(=\frac{1}{243{{x}^{5}}}\left[ 1+(-5)\left( \frac{2}{3x} \right)+\frac{(-5)(-6)}{2!}{{\left( \frac{2}{3x} \right)}^{2}}+\frac{(-5)(-6)(-7)}{3!}{{\left( \frac{2}{3x} \right)}^{3}}+…. \right]\).

\(=\frac{1}{243{{x}^{5}}}\left[ 1-\frac{10}{3x}+\frac{20}{3{{x}^{2}}}-\frac{280}{27{{x}^{3}}}+…. \right]\).

\(=\frac{1}{243}\left[ \frac{1}{{{x}^{5}}}-\frac{10}{3x}.\frac{1}{{{x}^{5}}}+\frac{20}{3{{x}^{2}}}.\frac{1}{{{x}^{5}}}-\frac{280}{27{{x}^{3}}}.\frac{1}{{{x}^{5}}}+…. \right]\).

**Note:**

- If n is a whole number, then there is no need of condition |x|<1
- In
the expansion (1 + x)
^{n}

(i) If n is natural number, then there are finite number terms exist.

(ii) If n is negative/ rational number, then there are infinite number of terms exist.

]]>**Important Dates:**

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Guru Gobind Singh Indraprastha University (GGSIPU) is first University established in 1998 by Govt. of NCT of Delhi under the provisions of Guru Gobind Singh Indraprastha University Act, 1998 read with its Amendment in 1999 The University is recognized by University Grants Commission (UGC), India under section 12B of UGC Act.

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]]>The All India Institute of
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]]>Binomial theorem: If |x| < 1 and p, q are positive integers then

**⇒ **\({{(1+x)}^{p/q}}=1+\frac{p}{1!}\left( \frac{x}{q} \right)+\frac{p(p-q)}{1.2}{{\left( \frac{x}{q} \right)}^{2}}+…+\frac{p(p-q)…(p-(r-1)q)}{1.2….r}{{\left( \frac{x}{q} \right)}^{r}}\).

**⇒ **\( {{(1-x)}^{p/q}}=1-\frac{p}{1!}\left( \frac{x}{q} \right)+\frac{p(p-q)}{1.2}{{\left( \frac{x}{q} \right)}^{2}}-…+{{(-1)}^{r}}\frac{p(p-q)…(p-(r-1)q)}{1.2….r}{{\left( \frac{x}{q} \right)}^{r}}\).

**⇒ **\( {{(1-x)}^{-p/q}}=1+\frac{p}{1!}\left( \frac{x}{q} \right)+\frac{p(p+q)}{1.2}{{\left( \frac{x}{q} \right)}^{2}}+…+\frac{p(p+q)…(p+(r-1)q)}{1.2….r}{{\left( \frac{x}{q} \right)}^{r}}\).

**⇒ **\( {{(1+x)}^{-p/q}}=1-\frac{p}{1!}\left( \frac{x}{q} \right)+\frac{p(p+q)}{1.2}{{\left( \frac{x}{q} \right)}^{2}}-…+{{(-1)}^{r}}\frac{p(p+q)…(p+(r-1)q)}{1.2….r}{{\left( \frac{x}{q} \right)}^{r}}\).

Obtain the values of x for which the binomial expansions for the follows

**Examples 1: **(6 + x)^{3/2}

**Solution: **Given that (6 + x)^{3/2}

= \( {{\left( 6 \right)}^{3/2}}{{\left( 1+\frac{x}{6} \right)}^{3/2}}\),

Hence, the binomial expansion for (6 + x)^{3/2} is
valid when |x/6|<1 ⇒
|x| < 6.

⇒ – 6 < x < 6

⇒ x ϵ (-6, 6)

**Examples 2:** (2 + 3x)⁻^{⅔}

**Solution: **Given that (2 + 3x)⁻^{⅔}

= \( {{\left( 2 \right)}^{-2/3}}{{\left( 1+\frac{3x}{2} \right)}^{-2/3}}\),

Hence, the binomial expansion for (2 + 3x)⁻^{⅔} is
valid when |3x/2| < 1.

⇒ |x| < ⅔

⇒ – ⅔ < x < ⅔

⇒ x ϵ (-⅔, ⅔).

]]>A circuit diagram can be defined as the graphical representation of an electrical circuit. An electrical circuit is a path or line through which an electrical current flow. The path may be closed, making it a loop. A closed circuit makes electrical current flow possible. It may also be an open circuit, where the electron flow is cut short because the path is broken. An open circuit doesn’t allow electrical current to flow. A schematic diagram shows the components of the circuit using standardized symbols.

**Simple Circuit Diagram:**

A circuit diagram is a simplified representation of the components of an electrical circuit. It shows the relative positions of all the elements and their connections to one another. The presentation of the interconnections between circuit components in the schematic diagram does not necessarily correspond to the physical arrangements in the finished device.

**Components of Circuit Diagram:**

**Ammeter: **It is used to measure the current passing at a particular point.

**Voltmeter: **It is used to measure the voltage between two points in a circuit.

**Battery: **It is a combination of cells.

**Switch: **It is a plug key and it is used to allow or stop the flow of current upon being pressed.

**Electric Bulb: **It is an electronic device which uses electricity to glow.

BITSAT-2019 is a Computer based online test for Admissions to Integrated First Degree Programmes of BITS Pilani Campuses in Pilani, Goa and Hyderabad. Due to large number of requests from the aspirants and parents, the last date to apply online for BITSAT-2019 along with application fee is extended till **5.00 PM, 31**^{st}** March, 2019**.

Mod Amplitude Form (or) Polar Form: Let z = a+ ib be a complex number such that |z| = r and θ be the amplitude of z. then cosθ = a/ r, sinθ = b/ r.

Now z = a + ib = r cosθ + i r sinθ

= r (cosθ + isinθ)

This know as mod (Modulus) amplitude form or polar form of z

**Key point:**

**⇒ **Cosθ + I sinθ is simply denoted by cisθ

**⇒ **Cosθ + isinθ = e^{iθ} is known as Euler’s formula

**⇒ **r₁ cisθ₁ = r₂ cisθ₂ ⇔ r₁ = r₂, θ₁ = 2kπ + θ₂, k ϵ Z

**⇒ **If z₁ = r₁ cisθ₁, z₂ = cisθ₂ then

⇝ z₁z₂ = cis (θ₁ + θ₂)

⇝ z₁/z₂ = r₁/r₂ cis (θ₁ – θ₂)

**Example:** \({{(\sqrt{3}+i)}^{100}}={{2}^{99}}(a+ib)\),
show that a² + b² = 4.

**Solution:** Given that \({{(\sqrt{3}+i)}^{100}}={{2}^{99}}(a+ib)\),

\(|{{(\sqrt{3}+i)}^{100}}|=|{{2}^{99}}(a+ib)|\),

\({{2}^{100}}={{2}^{99}}\sqrt{{{a}^{2}}+{{b}^{2}}}\),

\(\frac{{{2}^{100}}}{{{2}^{99}}}=\sqrt{{{a}^{2}}+{{b}^{2}}}\),

\(2=\sqrt{{{a}^{2}}+{{b}^{2}}}\),

Squaring on both sides

a² + b² = 4.

]]>